2
$\begingroup$

I have to prove the inequality $$ \frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy} \geq \frac{3}{1+\frac{(x+y)^2}{4}} $$ when $x^2+y^2=1$, using Cauchy-Schwarz Inequality.

The RHS is equal to $\frac{12}{5+2xy}$. I can prove, using C-S, that the RHS is $\geq \frac{9}{4+xy}$ or $\geq \frac{12}{5+4xy}$ but I can't go further.


I can prove the inequality only using A.M.-G.M. inequality proving that $xy\leq\frac{1}{2}$ and simplifying the expression all together, but this is not what I want.

Thanks

$\endgroup$
4
$\begingroup$

For $xy\geq0$ by C-S $$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}=\frac{3}{2+x^2y^2}+\frac{1}{1+xy}=$$ $$=\frac{4}{\frac{4(2+x^2y^2)}{3}}+\frac{1}{1+xy}\geq\frac{(2+1)^2}{\frac{4(2+x^2y^2)}{3}+1+xy}.$$ Id est, it's enough to prove that $$\frac{9}{\frac{4(2+x^2y^2)}{3}+1+xy}\geq\frac{12}{5+2xy}$$ or $$(1-2xy)(1+8xy)\geq0,$$ which is true by C-S again: $$2=(x^2+y^2)(1^2+1^2)\geq(x+y)^2=1+2xy,$$ which gives $1-2xy\geq0.$

For $xy\leq0$ by C-S again we obtain $$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}=\frac{1}{\frac{2+x^2y^2}{3}}+\frac{1}{1+xy}\geq\frac{4}{\frac{2+x^2y^2}{3}+1+xy}.$$ Thus, it's enough to prove that $$\frac{4}{\frac{2+x^2y^2}{3}+1+xy}\geq\frac{12}{5+2xy}$$ or $$x^2y^2+xy-2\leq0$$ or $$(1-xy)(2+xy)\geq0,$$ which is obvious.

$\endgroup$
  • $\begingroup$ Thanks. I like very much the last step! In your opinion is it possible to prove the inequality without rearranging the LHS and RHS together? $\endgroup$ – Alex Apr 17 at 7:43
  • $\begingroup$ I think, we can't because $-\frac{12}{5+2xy}$ is negative. $\endgroup$ – Michael Rozenberg Apr 17 at 7:49
  • 1
    $\begingroup$ Could you explain the first inequality further? I don't understand how to deduce it from the Cauchy-Schwartz. Thanks in advance. $\endgroup$ – awllower Apr 17 at 7:51
  • 1
    $\begingroup$ @MichaelRozenberg I found it as an exercise on a chapter on C-S. I don't know! Thaks:) $\endgroup$ – Alex Apr 17 at 8:05
  • 1
    $\begingroup$ @awllower You are right! My proof works for non-negative variables. I see a way, how we can fix it, but today I am very very busy. I'll fix it later. Thank you! $\endgroup$ – Michael Rozenberg Apr 18 at 7:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.