Proving an expression is non-negative on the interval $(-1,1)$
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I'm having trouble proving that the following expression is non-negative for $alpha^2<1$ and $theta > 0$:
$$frac{1}{(1-alpha)^2}+frac{1}{(1+alpha)^2}+frac{1}{(1-alpha+theta)^2}+frac{1}{(1+alpha+theta)^2}-frac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}$$
Here is a Mathematica plot of this expression for $theta = frac{1}{2}$:
I've tried to find a lower bound on the expression that is always non-negative, but any bound I find turns out to be negative in some places.
algebra-precalculus inequality upper-lower-bounds
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I'm having trouble proving that the following expression is non-negative for $alpha^2<1$ and $theta > 0$:
$$frac{1}{(1-alpha)^2}+frac{1}{(1+alpha)^2}+frac{1}{(1-alpha+theta)^2}+frac{1}{(1+alpha+theta)^2}-frac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}$$
Here is a Mathematica plot of this expression for $theta = frac{1}{2}$:
I've tried to find a lower bound on the expression that is always non-negative, but any bound I find turns out to be negative in some places.
algebra-precalculus inequality upper-lower-bounds
We can say $dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}$
– Yadati Kiran
Nov 22 at 15:24
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I'm having trouble proving that the following expression is non-negative for $alpha^2<1$ and $theta > 0$:
$$frac{1}{(1-alpha)^2}+frac{1}{(1+alpha)^2}+frac{1}{(1-alpha+theta)^2}+frac{1}{(1+alpha+theta)^2}-frac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}$$
Here is a Mathematica plot of this expression for $theta = frac{1}{2}$:
I've tried to find a lower bound on the expression that is always non-negative, but any bound I find turns out to be negative in some places.
algebra-precalculus inequality upper-lower-bounds
I'm having trouble proving that the following expression is non-negative for $alpha^2<1$ and $theta > 0$:
$$frac{1}{(1-alpha)^2}+frac{1}{(1+alpha)^2}+frac{1}{(1-alpha+theta)^2}+frac{1}{(1+alpha+theta)^2}-frac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}$$
Here is a Mathematica plot of this expression for $theta = frac{1}{2}$:
I've tried to find a lower bound on the expression that is always non-negative, but any bound I find turns out to be negative in some places.
algebra-precalculus inequality upper-lower-bounds
algebra-precalculus inequality upper-lower-bounds
asked Nov 22 at 14:55
user76284
1,1671124
1,1671124
We can say $dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}$
– Yadati Kiran
Nov 22 at 15:24
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We can say $dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}$
– Yadati Kiran
Nov 22 at 15:24
We can say $dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}$
– Yadati Kiran
Nov 22 at 15:24
We can say $dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}$
– Yadati Kiran
Nov 22 at 15:24
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$dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=2left(dfrac{2(1+alpha^2+theta^2+2theta)}{(1-alpha+theta)^2(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=4left(dfrac{(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2}{(1-alpha^2+theta)^2(1+alpha+theta)^2}right)>0$
The last part requires some calculation but it can be shown that $(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2$ does not have any roots for $alpha^2<1$ and $theta>0$.
Thanks, but how would you show that expression is non-negative? It seems comparable in difficulty to the original problem.
– user76284
Nov 22 at 18:29
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1 Answer
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$dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=2left(dfrac{2(1+alpha^2+theta^2+2theta)}{(1-alpha+theta)^2(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=4left(dfrac{(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2}{(1-alpha^2+theta)^2(1+alpha+theta)^2}right)>0$
The last part requires some calculation but it can be shown that $(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2$ does not have any roots for $alpha^2<1$ and $theta>0$.
Thanks, but how would you show that expression is non-negative? It seems comparable in difficulty to the original problem.
– user76284
Nov 22 at 18:29
add a comment |
up vote
0
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$dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=2left(dfrac{2(1+alpha^2+theta^2+2theta)}{(1-alpha+theta)^2(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=4left(dfrac{(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2}{(1-alpha^2+theta)^2(1+alpha+theta)^2}right)>0$
The last part requires some calculation but it can be shown that $(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2$ does not have any roots for $alpha^2<1$ and $theta>0$.
Thanks, but how would you show that expression is non-negative? It seems comparable in difficulty to the original problem.
– user76284
Nov 22 at 18:29
add a comment |
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0
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$dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=2left(dfrac{2(1+alpha^2+theta^2+2theta)}{(1-alpha+theta)^2(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=4left(dfrac{(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2}{(1-alpha^2+theta)^2(1+alpha+theta)^2}right)>0$
The last part requires some calculation but it can be shown that $(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2$ does not have any roots for $alpha^2<1$ and $theta>0$.
$dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=2left(dfrac{2(1+alpha^2+theta^2+2theta)}{(1-alpha+theta)^2(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=4left(dfrac{(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2}{(1-alpha^2+theta)^2(1+alpha+theta)^2}right)>0$
The last part requires some calculation but it can be shown that $(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2$ does not have any roots for $alpha^2<1$ and $theta>0$.
answered Nov 22 at 15:35
Yadati Kiran
1,352418
1,352418
Thanks, but how would you show that expression is non-negative? It seems comparable in difficulty to the original problem.
– user76284
Nov 22 at 18:29
add a comment |
Thanks, but how would you show that expression is non-negative? It seems comparable in difficulty to the original problem.
– user76284
Nov 22 at 18:29
Thanks, but how would you show that expression is non-negative? It seems comparable in difficulty to the original problem.
– user76284
Nov 22 at 18:29
Thanks, but how would you show that expression is non-negative? It seems comparable in difficulty to the original problem.
– user76284
Nov 22 at 18:29
add a comment |
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We can say $dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}$
– Yadati Kiran
Nov 22 at 15:24