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 $theta = frac{1}{2}$:



enter image description here



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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  • 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 $theta = frac{1}{2}$:



enter image description here



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







share|cite|improve this question






















  • 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













up vote
0
down vote

favorite









up vote
0
down vote

favorite











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 $theta = frac{1}{2}$:



enter image description here



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







share|cite|improve this question













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 $theta = frac{1}{2}$:



enter image description here



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






share|cite|improve this question













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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


















  • 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 $theta>0$.






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  • 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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$dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=2left(dfrac{2(1+alpha^2+theta^2+2theta)}{(1-alpha+theta)^2(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=4left(dfrac{(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2}{(1-alpha^2+theta)^2(1+alpha+theta)^2}right)>0$



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 $theta>0$.






share|cite|improve this answer





















  • 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















up vote
0
down vote













$dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=2left(dfrac{2(1+alpha^2+theta^2+2theta)}{(1-alpha+theta)^2(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=4left(dfrac{(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2}{(1-alpha^2+theta)^2(1+alpha+theta)^2}right)>0$



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 $theta>0$.






share|cite|improve this answer





















  • 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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$dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=2left(dfrac{2(1+alpha^2+theta^2+2theta)}{(1-alpha+theta)^2(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=4left(dfrac{(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2}{(1-alpha^2+theta)^2(1+alpha+theta)^2}right)>0$



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 $theta>0$.






share|cite|improve this answer












$dfrac{1}{(1-alpha)^2}+dfrac{1}{(1+alpha)^2}+dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}>2left(dfrac{1}{(1-alpha+theta)^2}+dfrac{1}{(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=2left(dfrac{2(1+alpha^2+theta^2+2theta)}{(1-alpha+theta)^2(1+alpha+theta)^2}right)-dfrac{4(1+alpha^2+theta)}{(1-alpha^2+theta)^2}=4left(dfrac{(1+alpha^2+theta^2+2theta)(1-alpha^2+theta)^2-(1+alpha^2+theta)(1+alpha+theta)^2}{(1-alpha^2+theta)^2(1+alpha+theta)^2}right)>0$



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 $theta>0$.







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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


















  • 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


















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