Find the maximum value of the expression ${frac {x}{1+x^2}} + {frac {y}{1+y^2}}+{frac {z}{1+z^2}}$....
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This question already has an answer here:
Prove that $frac{a}{a^2+1}+frac{b}{b^2+1}+frac{c}{c^2+1} leq frac{9}{10}$ if $a+b+c=1$.
2 answers
Find the maximum value of the expression :
$${frac {x}{1+x^2}} + {frac {y}{1+y^2}}+{frac {z}{1+z^2}}$$
where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$.
Taking $x=y=z=frac {1}{3}$, I get the expression as $frac {3x}{1+x^2}$, which is equal to $frac {1}{1+{frac{1}{9}}}$ or $frac {9}{10}$.
How can I actually solve the problem without making unnecessary assumptions ?
inequality maxima-minima uvw
marked as duplicate by Nosrati, Davide Giraudo, Rebellos, José Carlos Santos, amWhy Nov 22 at 19:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
5
down vote
favorite
This question already has an answer here:
Prove that $frac{a}{a^2+1}+frac{b}{b^2+1}+frac{c}{c^2+1} leq frac{9}{10}$ if $a+b+c=1$.
2 answers
Find the maximum value of the expression :
$${frac {x}{1+x^2}} + {frac {y}{1+y^2}}+{frac {z}{1+z^2}}$$
where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$.
Taking $x=y=z=frac {1}{3}$, I get the expression as $frac {3x}{1+x^2}$, which is equal to $frac {1}{1+{frac{1}{9}}}$ or $frac {9}{10}$.
How can I actually solve the problem without making unnecessary assumptions ?
inequality maxima-minima uvw
marked as duplicate by Nosrati, Davide Giraudo, Rebellos, José Carlos Santos, amWhy Nov 22 at 19:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Have you tried Lagrange multipliers ? You can also substitute $z=1-x-y$.
– Yves Daoust
Jan 13 '17 at 7:30
$x=cot A$ etc.
– lab bhattacharjee
Jan 13 '17 at 7:32
Oooops, with $z=1-x-y$, the computation is terrible.
– Yves Daoust
Jan 13 '17 at 7:41
This inequality was in Poland MO 1996.
– Michael Rozenberg
Oct 25 at 13:28
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
This question already has an answer here:
Prove that $frac{a}{a^2+1}+frac{b}{b^2+1}+frac{c}{c^2+1} leq frac{9}{10}$ if $a+b+c=1$.
2 answers
Find the maximum value of the expression :
$${frac {x}{1+x^2}} + {frac {y}{1+y^2}}+{frac {z}{1+z^2}}$$
where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$.
Taking $x=y=z=frac {1}{3}$, I get the expression as $frac {3x}{1+x^2}$, which is equal to $frac {1}{1+{frac{1}{9}}}$ or $frac {9}{10}$.
How can I actually solve the problem without making unnecessary assumptions ?
inequality maxima-minima uvw
This question already has an answer here:
Prove that $frac{a}{a^2+1}+frac{b}{b^2+1}+frac{c}{c^2+1} leq frac{9}{10}$ if $a+b+c=1$.
2 answers
Find the maximum value of the expression :
$${frac {x}{1+x^2}} + {frac {y}{1+y^2}}+{frac {z}{1+z^2}}$$
where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$.
Taking $x=y=z=frac {1}{3}$, I get the expression as $frac {3x}{1+x^2}$, which is equal to $frac {1}{1+{frac{1}{9}}}$ or $frac {9}{10}$.
How can I actually solve the problem without making unnecessary assumptions ?
This question already has an answer here:
Prove that $frac{a}{a^2+1}+frac{b}{b^2+1}+frac{c}{c^2+1} leq frac{9}{10}$ if $a+b+c=1$.
2 answers
inequality maxima-minima uvw
inequality maxima-minima uvw
edited Nov 22 at 14:45
Nosrati
26.3k62353
26.3k62353
asked Jan 13 '17 at 7:26
Nirbhay
1,270427
1,270427
marked as duplicate by Nosrati, Davide Giraudo, Rebellos, José Carlos Santos, amWhy Nov 22 at 19:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Nosrati, Davide Giraudo, Rebellos, José Carlos Santos, amWhy Nov 22 at 19:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Have you tried Lagrange multipliers ? You can also substitute $z=1-x-y$.
– Yves Daoust
Jan 13 '17 at 7:30
$x=cot A$ etc.
– lab bhattacharjee
Jan 13 '17 at 7:32
Oooops, with $z=1-x-y$, the computation is terrible.
– Yves Daoust
Jan 13 '17 at 7:41
This inequality was in Poland MO 1996.
– Michael Rozenberg
Oct 25 at 13:28
add a comment |
1
Have you tried Lagrange multipliers ? You can also substitute $z=1-x-y$.
– Yves Daoust
Jan 13 '17 at 7:30
$x=cot A$ etc.
– lab bhattacharjee
Jan 13 '17 at 7:32
Oooops, with $z=1-x-y$, the computation is terrible.
– Yves Daoust
Jan 13 '17 at 7:41
This inequality was in Poland MO 1996.
– Michael Rozenberg
Oct 25 at 13:28
1
1
Have you tried Lagrange multipliers ? You can also substitute $z=1-x-y$.
– Yves Daoust
Jan 13 '17 at 7:30
Have you tried Lagrange multipliers ? You can also substitute $z=1-x-y$.
– Yves Daoust
Jan 13 '17 at 7:30
$x=cot A$ etc.
– lab bhattacharjee
Jan 13 '17 at 7:32
$x=cot A$ etc.
– lab bhattacharjee
Jan 13 '17 at 7:32
Oooops, with $z=1-x-y$, the computation is terrible.
– Yves Daoust
Jan 13 '17 at 7:41
Oooops, with $z=1-x-y$, the computation is terrible.
– Yves Daoust
Jan 13 '17 at 7:41
This inequality was in Poland MO 1996.
– Michael Rozenberg
Oct 25 at 13:28
This inequality was in Poland MO 1996.
– Michael Rozenberg
Oct 25 at 13:28
add a comment |
4 Answers
4
active
oldest
votes
up vote
5
down vote
accepted
We use the method of Lagrange Multipliers.
First, we define the function to maximize as $f(x,y,z)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}$ and our constraint as $g(x,y,z)=x+y+z-1$.
Thus, the Lagrange function is:
$$mathcal{L}(x,y,z,lambda)=f(x,y,z)+lambdacdot g(x,y,z)$$
$$mathcal{L}(x,y,z,lambda)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}+lambda(x+y+z-1) tag{1}$$
Where $lambda$ is some scalar. This function's partial derivatives must equal all equal to zero:
$$frac{partial mathcal{L}}{partial x}=frac{1-x^2}{(x^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial y}=frac{1-y^2}{(y^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial z}=frac{1-z^2}{(z^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial lambda}=x+y+z-1=0$$
Solving this system of equations will yield values for $x,y,z,lambda$ which will either minimize or maximize $f(x,y,z)$.
After some tedious calculation, we notice that the solution which maximizes the value of $f(x,y,z)$ is:
$x=y=z=frac{1}{3}$ and $lambda=-0.72$, as you have correctly predicted.
Thus, the maximum value of $f(x,y,z)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}=frac{9}{10}$.
1
Thanks for the explanation :) +1
– Nirbhay
Jan 13 '17 at 9:49
add a comment |
up vote
2
down vote
Let $x=frac{a}{3}$, $y=frac{b}{3}$ and $z=frac{c}{3}$.
Hence, $a+b+c=3$ and we need to prove that $$sum_{cyc}frac{a}{9+a^2}leqfrac{3}{10}$$ or
$$sum_{cyc}left(frac{1}{10}-frac{a}{9+a^2}right)geq0$$ or
$$sum_{cyc}frac{a^2-10a+9}{9+a^2}geq0$$ or
$$sum_{cyc}left(frac{a^2-10a+9}{9+a^2}+frac{2}{3}right)geq2$$ or
$$sum_{cyc}frac{(a-3)^2}{9+a^2}geqfrac{6}{5}.$$
Let $k=3sqrt2-6$. Hence, by C-S
$$sum_{cyc}frac{(a-3)^2}{9+a^2}=sum_{cyc}frac{(a-3)^2(a+k)^2}{(a+k)^2(9+a^2)}geqfrac{left(sumlimits_{cyc}(a-3)(a+k)right)^2}{sumlimits_{cyc}(a+k)^2(a^2+9)}=frac{left(sumlimits_{cyc}(a^2-3-2k)right)^2}{sumlimits_{cyc}(a+k)^2(a^2+9)}.$$
Thus, it remains to prove that
$$5left(sumlimits_{cyc}(a^2-3-2k)right)^2geq6sumlimits_{cyc}(a+k)^2(a^2+9),$$
which is fourth degree.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative and $abc=w^3$.
Hence, the last inequality is a linear inequality of $w^3$, which says that
it's enough to prove the last inequality for en extremal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$. We obtain
$$(a-1)^2(sqrt2a-3sqrt2+3)^2geq0.$$
Done!
That is very cool approach (+unity) but at the same time very tedious....
– Nirbhay
Feb 6 '17 at 11:13
1
@Nirbhay I have another solution by uvw. I am ready to show.
– Michael Rozenberg
Feb 6 '17 at 14:55
Go ahead !!! :) By the way I want to ask you (since you are so good at ineqs), which techniques are the most helpful while solving cyclic ineqs ??? Is it necessary to learn calculus for ineqs(for eg. while using Jensen's ineq) ??? In what order should I learn the various ineqs ???
– Nirbhay
Feb 6 '17 at 15:12
1
@Nirbhay I'll show my proof in evening. I think if you wish yo be good in proving of cyclic inequalities, you need just to prove very many inequalities or to see, how humans do it. For which see posts of Vasc, Can, qa and more.
– Michael Rozenberg
Feb 6 '17 at 15:47
It is almost "Good Night" for me here LOL XD
– Nirbhay
Feb 6 '17 at 16:00
add a comment |
up vote
2
down vote
Let $x+y+z=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
We need to prove that $$sum_{cyc}frac{x}{x^2+(x+y+z)^2}leqfrac{9}{10(x+y+z)}$$ or
$$9prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)geq0,$$
which is $9w^6+A(u,v^2)w^3+B(u,v^2)geq0$, which says that
$$9prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)geq$$
$$geqfrac{1}{81}left(sum_{cyc}(4x^3-3x^2y-3x^2z+2xyz)right)^2$$
is a linear inequality of $w^3$.
Thus, by $uvw$ it remains to prove the last inequality for $y=z=1$, which gives
$$(x-1)^2(x+2)^2(79x^2+253x+253)geq0,$$
which is obviously true.
Done!
add a comment |
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0
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Threw come to mind.
The first is Lagrange Multipliers.
The second is to reason directly. Plus in $(x,y,z)$ and then plug in $(x-a,y+b,z+c)$ where $a+b+c=0$. You can use algebra to show that the former produces a larger total than the latter, and it follows that the maximum occurs when all the values are equal.
The third is that by symmetry. First observe that the max must have non-negative coordinates. That condition, plus $x+y+zleq 1$ define a convex optimization problem, whose maximum must be at a corner of the convex hull. It's simple to check that of the corners, $x=y=z$ gives the largest value.
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
We use the method of Lagrange Multipliers.
First, we define the function to maximize as $f(x,y,z)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}$ and our constraint as $g(x,y,z)=x+y+z-1$.
Thus, the Lagrange function is:
$$mathcal{L}(x,y,z,lambda)=f(x,y,z)+lambdacdot g(x,y,z)$$
$$mathcal{L}(x,y,z,lambda)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}+lambda(x+y+z-1) tag{1}$$
Where $lambda$ is some scalar. This function's partial derivatives must equal all equal to zero:
$$frac{partial mathcal{L}}{partial x}=frac{1-x^2}{(x^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial y}=frac{1-y^2}{(y^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial z}=frac{1-z^2}{(z^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial lambda}=x+y+z-1=0$$
Solving this system of equations will yield values for $x,y,z,lambda$ which will either minimize or maximize $f(x,y,z)$.
After some tedious calculation, we notice that the solution which maximizes the value of $f(x,y,z)$ is:
$x=y=z=frac{1}{3}$ and $lambda=-0.72$, as you have correctly predicted.
Thus, the maximum value of $f(x,y,z)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}=frac{9}{10}$.
1
Thanks for the explanation :) +1
– Nirbhay
Jan 13 '17 at 9:49
add a comment |
up vote
5
down vote
accepted
We use the method of Lagrange Multipliers.
First, we define the function to maximize as $f(x,y,z)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}$ and our constraint as $g(x,y,z)=x+y+z-1$.
Thus, the Lagrange function is:
$$mathcal{L}(x,y,z,lambda)=f(x,y,z)+lambdacdot g(x,y,z)$$
$$mathcal{L}(x,y,z,lambda)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}+lambda(x+y+z-1) tag{1}$$
Where $lambda$ is some scalar. This function's partial derivatives must equal all equal to zero:
$$frac{partial mathcal{L}}{partial x}=frac{1-x^2}{(x^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial y}=frac{1-y^2}{(y^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial z}=frac{1-z^2}{(z^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial lambda}=x+y+z-1=0$$
Solving this system of equations will yield values for $x,y,z,lambda$ which will either minimize or maximize $f(x,y,z)$.
After some tedious calculation, we notice that the solution which maximizes the value of $f(x,y,z)$ is:
$x=y=z=frac{1}{3}$ and $lambda=-0.72$, as you have correctly predicted.
Thus, the maximum value of $f(x,y,z)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}=frac{9}{10}$.
1
Thanks for the explanation :) +1
– Nirbhay
Jan 13 '17 at 9:49
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
We use the method of Lagrange Multipliers.
First, we define the function to maximize as $f(x,y,z)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}$ and our constraint as $g(x,y,z)=x+y+z-1$.
Thus, the Lagrange function is:
$$mathcal{L}(x,y,z,lambda)=f(x,y,z)+lambdacdot g(x,y,z)$$
$$mathcal{L}(x,y,z,lambda)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}+lambda(x+y+z-1) tag{1}$$
Where $lambda$ is some scalar. This function's partial derivatives must equal all equal to zero:
$$frac{partial mathcal{L}}{partial x}=frac{1-x^2}{(x^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial y}=frac{1-y^2}{(y^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial z}=frac{1-z^2}{(z^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial lambda}=x+y+z-1=0$$
Solving this system of equations will yield values for $x,y,z,lambda$ which will either minimize or maximize $f(x,y,z)$.
After some tedious calculation, we notice that the solution which maximizes the value of $f(x,y,z)$ is:
$x=y=z=frac{1}{3}$ and $lambda=-0.72$, as you have correctly predicted.
Thus, the maximum value of $f(x,y,z)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}=frac{9}{10}$.
We use the method of Lagrange Multipliers.
First, we define the function to maximize as $f(x,y,z)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}$ and our constraint as $g(x,y,z)=x+y+z-1$.
Thus, the Lagrange function is:
$$mathcal{L}(x,y,z,lambda)=f(x,y,z)+lambdacdot g(x,y,z)$$
$$mathcal{L}(x,y,z,lambda)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}+lambda(x+y+z-1) tag{1}$$
Where $lambda$ is some scalar. This function's partial derivatives must equal all equal to zero:
$$frac{partial mathcal{L}}{partial x}=frac{1-x^2}{(x^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial y}=frac{1-y^2}{(y^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial z}=frac{1-z^2}{(z^2+1)^2}+lambda=0$$
$$frac{partial mathcal{L}}{partial lambda}=x+y+z-1=0$$
Solving this system of equations will yield values for $x,y,z,lambda$ which will either minimize or maximize $f(x,y,z)$.
After some tedious calculation, we notice that the solution which maximizes the value of $f(x,y,z)$ is:
$x=y=z=frac{1}{3}$ and $lambda=-0.72$, as you have correctly predicted.
Thus, the maximum value of $f(x,y,z)=frac{x}{1+x^2}+frac{y}{1+y^2}+frac{z}{1+z^2}=frac{9}{10}$.
answered Jan 13 '17 at 9:47
projectilemotion
11.4k62041
11.4k62041
1
Thanks for the explanation :) +1
– Nirbhay
Jan 13 '17 at 9:49
add a comment |
1
Thanks for the explanation :) +1
– Nirbhay
Jan 13 '17 at 9:49
1
1
Thanks for the explanation :) +1
– Nirbhay
Jan 13 '17 at 9:49
Thanks for the explanation :) +1
– Nirbhay
Jan 13 '17 at 9:49
add a comment |
up vote
2
down vote
Let $x=frac{a}{3}$, $y=frac{b}{3}$ and $z=frac{c}{3}$.
Hence, $a+b+c=3$ and we need to prove that $$sum_{cyc}frac{a}{9+a^2}leqfrac{3}{10}$$ or
$$sum_{cyc}left(frac{1}{10}-frac{a}{9+a^2}right)geq0$$ or
$$sum_{cyc}frac{a^2-10a+9}{9+a^2}geq0$$ or
$$sum_{cyc}left(frac{a^2-10a+9}{9+a^2}+frac{2}{3}right)geq2$$ or
$$sum_{cyc}frac{(a-3)^2}{9+a^2}geqfrac{6}{5}.$$
Let $k=3sqrt2-6$. Hence, by C-S
$$sum_{cyc}frac{(a-3)^2}{9+a^2}=sum_{cyc}frac{(a-3)^2(a+k)^2}{(a+k)^2(9+a^2)}geqfrac{left(sumlimits_{cyc}(a-3)(a+k)right)^2}{sumlimits_{cyc}(a+k)^2(a^2+9)}=frac{left(sumlimits_{cyc}(a^2-3-2k)right)^2}{sumlimits_{cyc}(a+k)^2(a^2+9)}.$$
Thus, it remains to prove that
$$5left(sumlimits_{cyc}(a^2-3-2k)right)^2geq6sumlimits_{cyc}(a+k)^2(a^2+9),$$
which is fourth degree.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative and $abc=w^3$.
Hence, the last inequality is a linear inequality of $w^3$, which says that
it's enough to prove the last inequality for en extremal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$. We obtain
$$(a-1)^2(sqrt2a-3sqrt2+3)^2geq0.$$
Done!
That is very cool approach (+unity) but at the same time very tedious....
– Nirbhay
Feb 6 '17 at 11:13
1
@Nirbhay I have another solution by uvw. I am ready to show.
– Michael Rozenberg
Feb 6 '17 at 14:55
Go ahead !!! :) By the way I want to ask you (since you are so good at ineqs), which techniques are the most helpful while solving cyclic ineqs ??? Is it necessary to learn calculus for ineqs(for eg. while using Jensen's ineq) ??? In what order should I learn the various ineqs ???
– Nirbhay
Feb 6 '17 at 15:12
1
@Nirbhay I'll show my proof in evening. I think if you wish yo be good in proving of cyclic inequalities, you need just to prove very many inequalities or to see, how humans do it. For which see posts of Vasc, Can, qa and more.
– Michael Rozenberg
Feb 6 '17 at 15:47
It is almost "Good Night" for me here LOL XD
– Nirbhay
Feb 6 '17 at 16:00
add a comment |
up vote
2
down vote
Let $x=frac{a}{3}$, $y=frac{b}{3}$ and $z=frac{c}{3}$.
Hence, $a+b+c=3$ and we need to prove that $$sum_{cyc}frac{a}{9+a^2}leqfrac{3}{10}$$ or
$$sum_{cyc}left(frac{1}{10}-frac{a}{9+a^2}right)geq0$$ or
$$sum_{cyc}frac{a^2-10a+9}{9+a^2}geq0$$ or
$$sum_{cyc}left(frac{a^2-10a+9}{9+a^2}+frac{2}{3}right)geq2$$ or
$$sum_{cyc}frac{(a-3)^2}{9+a^2}geqfrac{6}{5}.$$
Let $k=3sqrt2-6$. Hence, by C-S
$$sum_{cyc}frac{(a-3)^2}{9+a^2}=sum_{cyc}frac{(a-3)^2(a+k)^2}{(a+k)^2(9+a^2)}geqfrac{left(sumlimits_{cyc}(a-3)(a+k)right)^2}{sumlimits_{cyc}(a+k)^2(a^2+9)}=frac{left(sumlimits_{cyc}(a^2-3-2k)right)^2}{sumlimits_{cyc}(a+k)^2(a^2+9)}.$$
Thus, it remains to prove that
$$5left(sumlimits_{cyc}(a^2-3-2k)right)^2geq6sumlimits_{cyc}(a+k)^2(a^2+9),$$
which is fourth degree.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative and $abc=w^3$.
Hence, the last inequality is a linear inequality of $w^3$, which says that
it's enough to prove the last inequality for en extremal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$. We obtain
$$(a-1)^2(sqrt2a-3sqrt2+3)^2geq0.$$
Done!
That is very cool approach (+unity) but at the same time very tedious....
– Nirbhay
Feb 6 '17 at 11:13
1
@Nirbhay I have another solution by uvw. I am ready to show.
– Michael Rozenberg
Feb 6 '17 at 14:55
Go ahead !!! :) By the way I want to ask you (since you are so good at ineqs), which techniques are the most helpful while solving cyclic ineqs ??? Is it necessary to learn calculus for ineqs(for eg. while using Jensen's ineq) ??? In what order should I learn the various ineqs ???
– Nirbhay
Feb 6 '17 at 15:12
1
@Nirbhay I'll show my proof in evening. I think if you wish yo be good in proving of cyclic inequalities, you need just to prove very many inequalities or to see, how humans do it. For which see posts of Vasc, Can, qa and more.
– Michael Rozenberg
Feb 6 '17 at 15:47
It is almost "Good Night" for me here LOL XD
– Nirbhay
Feb 6 '17 at 16:00
add a comment |
up vote
2
down vote
up vote
2
down vote
Let $x=frac{a}{3}$, $y=frac{b}{3}$ and $z=frac{c}{3}$.
Hence, $a+b+c=3$ and we need to prove that $$sum_{cyc}frac{a}{9+a^2}leqfrac{3}{10}$$ or
$$sum_{cyc}left(frac{1}{10}-frac{a}{9+a^2}right)geq0$$ or
$$sum_{cyc}frac{a^2-10a+9}{9+a^2}geq0$$ or
$$sum_{cyc}left(frac{a^2-10a+9}{9+a^2}+frac{2}{3}right)geq2$$ or
$$sum_{cyc}frac{(a-3)^2}{9+a^2}geqfrac{6}{5}.$$
Let $k=3sqrt2-6$. Hence, by C-S
$$sum_{cyc}frac{(a-3)^2}{9+a^2}=sum_{cyc}frac{(a-3)^2(a+k)^2}{(a+k)^2(9+a^2)}geqfrac{left(sumlimits_{cyc}(a-3)(a+k)right)^2}{sumlimits_{cyc}(a+k)^2(a^2+9)}=frac{left(sumlimits_{cyc}(a^2-3-2k)right)^2}{sumlimits_{cyc}(a+k)^2(a^2+9)}.$$
Thus, it remains to prove that
$$5left(sumlimits_{cyc}(a^2-3-2k)right)^2geq6sumlimits_{cyc}(a+k)^2(a^2+9),$$
which is fourth degree.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative and $abc=w^3$.
Hence, the last inequality is a linear inequality of $w^3$, which says that
it's enough to prove the last inequality for en extremal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$. We obtain
$$(a-1)^2(sqrt2a-3sqrt2+3)^2geq0.$$
Done!
Let $x=frac{a}{3}$, $y=frac{b}{3}$ and $z=frac{c}{3}$.
Hence, $a+b+c=3$ and we need to prove that $$sum_{cyc}frac{a}{9+a^2}leqfrac{3}{10}$$ or
$$sum_{cyc}left(frac{1}{10}-frac{a}{9+a^2}right)geq0$$ or
$$sum_{cyc}frac{a^2-10a+9}{9+a^2}geq0$$ or
$$sum_{cyc}left(frac{a^2-10a+9}{9+a^2}+frac{2}{3}right)geq2$$ or
$$sum_{cyc}frac{(a-3)^2}{9+a^2}geqfrac{6}{5}.$$
Let $k=3sqrt2-6$. Hence, by C-S
$$sum_{cyc}frac{(a-3)^2}{9+a^2}=sum_{cyc}frac{(a-3)^2(a+k)^2}{(a+k)^2(9+a^2)}geqfrac{left(sumlimits_{cyc}(a-3)(a+k)right)^2}{sumlimits_{cyc}(a+k)^2(a^2+9)}=frac{left(sumlimits_{cyc}(a^2-3-2k)right)^2}{sumlimits_{cyc}(a+k)^2(a^2+9)}.$$
Thus, it remains to prove that
$$5left(sumlimits_{cyc}(a^2-3-2k)right)^2geq6sumlimits_{cyc}(a+k)^2(a^2+9),$$
which is fourth degree.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative and $abc=w^3$.
Hence, the last inequality is a linear inequality of $w^3$, which says that
it's enough to prove the last inequality for en extremal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$. We obtain
$$(a-1)^2(sqrt2a-3sqrt2+3)^2geq0.$$
Done!
answered Feb 5 '17 at 21:13
Michael Rozenberg
95k1588183
95k1588183
That is very cool approach (+unity) but at the same time very tedious....
– Nirbhay
Feb 6 '17 at 11:13
1
@Nirbhay I have another solution by uvw. I am ready to show.
– Michael Rozenberg
Feb 6 '17 at 14:55
Go ahead !!! :) By the way I want to ask you (since you are so good at ineqs), which techniques are the most helpful while solving cyclic ineqs ??? Is it necessary to learn calculus for ineqs(for eg. while using Jensen's ineq) ??? In what order should I learn the various ineqs ???
– Nirbhay
Feb 6 '17 at 15:12
1
@Nirbhay I'll show my proof in evening. I think if you wish yo be good in proving of cyclic inequalities, you need just to prove very many inequalities or to see, how humans do it. For which see posts of Vasc, Can, qa and more.
– Michael Rozenberg
Feb 6 '17 at 15:47
It is almost "Good Night" for me here LOL XD
– Nirbhay
Feb 6 '17 at 16:00
add a comment |
That is very cool approach (+unity) but at the same time very tedious....
– Nirbhay
Feb 6 '17 at 11:13
1
@Nirbhay I have another solution by uvw. I am ready to show.
– Michael Rozenberg
Feb 6 '17 at 14:55
Go ahead !!! :) By the way I want to ask you (since you are so good at ineqs), which techniques are the most helpful while solving cyclic ineqs ??? Is it necessary to learn calculus for ineqs(for eg. while using Jensen's ineq) ??? In what order should I learn the various ineqs ???
– Nirbhay
Feb 6 '17 at 15:12
1
@Nirbhay I'll show my proof in evening. I think if you wish yo be good in proving of cyclic inequalities, you need just to prove very many inequalities or to see, how humans do it. For which see posts of Vasc, Can, qa and more.
– Michael Rozenberg
Feb 6 '17 at 15:47
It is almost "Good Night" for me here LOL XD
– Nirbhay
Feb 6 '17 at 16:00
That is very cool approach (+unity) but at the same time very tedious....
– Nirbhay
Feb 6 '17 at 11:13
That is very cool approach (+unity) but at the same time very tedious....
– Nirbhay
Feb 6 '17 at 11:13
1
1
@Nirbhay I have another solution by uvw. I am ready to show.
– Michael Rozenberg
Feb 6 '17 at 14:55
@Nirbhay I have another solution by uvw. I am ready to show.
– Michael Rozenberg
Feb 6 '17 at 14:55
Go ahead !!! :) By the way I want to ask you (since you are so good at ineqs), which techniques are the most helpful while solving cyclic ineqs ??? Is it necessary to learn calculus for ineqs(for eg. while using Jensen's ineq) ??? In what order should I learn the various ineqs ???
– Nirbhay
Feb 6 '17 at 15:12
Go ahead !!! :) By the way I want to ask you (since you are so good at ineqs), which techniques are the most helpful while solving cyclic ineqs ??? Is it necessary to learn calculus for ineqs(for eg. while using Jensen's ineq) ??? In what order should I learn the various ineqs ???
– Nirbhay
Feb 6 '17 at 15:12
1
1
@Nirbhay I'll show my proof in evening. I think if you wish yo be good in proving of cyclic inequalities, you need just to prove very many inequalities or to see, how humans do it. For which see posts of Vasc, Can, qa and more.
– Michael Rozenberg
Feb 6 '17 at 15:47
@Nirbhay I'll show my proof in evening. I think if you wish yo be good in proving of cyclic inequalities, you need just to prove very many inequalities or to see, how humans do it. For which see posts of Vasc, Can, qa and more.
– Michael Rozenberg
Feb 6 '17 at 15:47
It is almost "Good Night" for me here LOL XD
– Nirbhay
Feb 6 '17 at 16:00
It is almost "Good Night" for me here LOL XD
– Nirbhay
Feb 6 '17 at 16:00
add a comment |
up vote
2
down vote
Let $x+y+z=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
We need to prove that $$sum_{cyc}frac{x}{x^2+(x+y+z)^2}leqfrac{9}{10(x+y+z)}$$ or
$$9prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)geq0,$$
which is $9w^6+A(u,v^2)w^3+B(u,v^2)geq0$, which says that
$$9prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)geq$$
$$geqfrac{1}{81}left(sum_{cyc}(4x^3-3x^2y-3x^2z+2xyz)right)^2$$
is a linear inequality of $w^3$.
Thus, by $uvw$ it remains to prove the last inequality for $y=z=1$, which gives
$$(x-1)^2(x+2)^2(79x^2+253x+253)geq0,$$
which is obviously true.
Done!
add a comment |
up vote
2
down vote
Let $x+y+z=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
We need to prove that $$sum_{cyc}frac{x}{x^2+(x+y+z)^2}leqfrac{9}{10(x+y+z)}$$ or
$$9prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)geq0,$$
which is $9w^6+A(u,v^2)w^3+B(u,v^2)geq0$, which says that
$$9prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)geq$$
$$geqfrac{1}{81}left(sum_{cyc}(4x^3-3x^2y-3x^2z+2xyz)right)^2$$
is a linear inequality of $w^3$.
Thus, by $uvw$ it remains to prove the last inequality for $y=z=1$, which gives
$$(x-1)^2(x+2)^2(79x^2+253x+253)geq0,$$
which is obviously true.
Done!
add a comment |
up vote
2
down vote
up vote
2
down vote
Let $x+y+z=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
We need to prove that $$sum_{cyc}frac{x}{x^2+(x+y+z)^2}leqfrac{9}{10(x+y+z)}$$ or
$$9prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)geq0,$$
which is $9w^6+A(u,v^2)w^3+B(u,v^2)geq0$, which says that
$$9prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)geq$$
$$geqfrac{1}{81}left(sum_{cyc}(4x^3-3x^2y-3x^2z+2xyz)right)^2$$
is a linear inequality of $w^3$.
Thus, by $uvw$ it remains to prove the last inequality for $y=z=1$, which gives
$$(x-1)^2(x+2)^2(79x^2+253x+253)geq0,$$
which is obviously true.
Done!
Let $x+y+z=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
We need to prove that $$sum_{cyc}frac{x}{x^2+(x+y+z)^2}leqfrac{9}{10(x+y+z)}$$ or
$$9prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)geq0,$$
which is $9w^6+A(u,v^2)w^3+B(u,v^2)geq0$, which says that
$$9prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)geq$$
$$geqfrac{1}{81}left(sum_{cyc}(4x^3-3x^2y-3x^2z+2xyz)right)^2$$
is a linear inequality of $w^3$.
Thus, by $uvw$ it remains to prove the last inequality for $y=z=1$, which gives
$$(x-1)^2(x+2)^2(79x^2+253x+253)geq0,$$
which is obviously true.
Done!
answered Feb 6 '17 at 18:50
Michael Rozenberg
95k1588183
95k1588183
add a comment |
add a comment |
up vote
0
down vote
Threw come to mind.
The first is Lagrange Multipliers.
The second is to reason directly. Plus in $(x,y,z)$ and then plug in $(x-a,y+b,z+c)$ where $a+b+c=0$. You can use algebra to show that the former produces a larger total than the latter, and it follows that the maximum occurs when all the values are equal.
The third is that by symmetry. First observe that the max must have non-negative coordinates. That condition, plus $x+y+zleq 1$ define a convex optimization problem, whose maximum must be at a corner of the convex hull. It's simple to check that of the corners, $x=y=z$ gives the largest value.
add a comment |
up vote
0
down vote
Threw come to mind.
The first is Lagrange Multipliers.
The second is to reason directly. Plus in $(x,y,z)$ and then plug in $(x-a,y+b,z+c)$ where $a+b+c=0$. You can use algebra to show that the former produces a larger total than the latter, and it follows that the maximum occurs when all the values are equal.
The third is that by symmetry. First observe that the max must have non-negative coordinates. That condition, plus $x+y+zleq 1$ define a convex optimization problem, whose maximum must be at a corner of the convex hull. It's simple to check that of the corners, $x=y=z$ gives the largest value.
add a comment |
up vote
0
down vote
up vote
0
down vote
Threw come to mind.
The first is Lagrange Multipliers.
The second is to reason directly. Plus in $(x,y,z)$ and then plug in $(x-a,y+b,z+c)$ where $a+b+c=0$. You can use algebra to show that the former produces a larger total than the latter, and it follows that the maximum occurs when all the values are equal.
The third is that by symmetry. First observe that the max must have non-negative coordinates. That condition, plus $x+y+zleq 1$ define a convex optimization problem, whose maximum must be at a corner of the convex hull. It's simple to check that of the corners, $x=y=z$ gives the largest value.
Threw come to mind.
The first is Lagrange Multipliers.
The second is to reason directly. Plus in $(x,y,z)$ and then plug in $(x-a,y+b,z+c)$ where $a+b+c=0$. You can use algebra to show that the former produces a larger total than the latter, and it follows that the maximum occurs when all the values are equal.
The third is that by symmetry. First observe that the max must have non-negative coordinates. That condition, plus $x+y+zleq 1$ define a convex optimization problem, whose maximum must be at a corner of the convex hull. It's simple to check that of the corners, $x=y=z$ gives the largest value.
answered Jan 13 '17 at 7:34
Stella Biderman
26.5k63175
26.5k63175
add a comment |
add a comment |
1
Have you tried Lagrange multipliers ? You can also substitute $z=1-x-y$.
– Yves Daoust
Jan 13 '17 at 7:30
$x=cot A$ etc.
– lab bhattacharjee
Jan 13 '17 at 7:32
Oooops, with $z=1-x-y$, the computation is terrible.
– Yves Daoust
Jan 13 '17 at 7:41
This inequality was in Poland MO 1996.
– Michael Rozenberg
Oct 25 at 13:28