2
$\begingroup$

For all $x,y>0$, $$\frac{1}{(x+1)^2} + \frac{1}{(y+1)^2} \ge \frac{1}{xy+1}$$

I can only think of substituting $x+1$ with $a$ and $y+1$ with $b$. Then the inequality turns into $$(a^2 + b^2) (ab-a-b +2) \ge a^2b^2$$

I can proceed no further. Please help.

$\endgroup$
1
$\begingroup$

After your substitutions we get new conditions $a>1$ and $b>1$, which makes the inequality harder.

By the way, it's just $$xy(x-y)^2+(xy-1)^2\geq0.$$ Also, we can use C-S: $$\sum_{cyc}\frac{1}{(x+1)^2}=\sum_{cyc}\frac{y}{(x\sqrt{y}+\sqrt{y})^2}\geq\sum_{cyc}\frac{y}{(xy+1)(x+y)}=\frac{1}{xy+1}.$$

$\endgroup$
  • $\begingroup$ Thank you! got it $\endgroup$ – ami_ba Sep 2 '18 at 6:26
  • $\begingroup$ You should replace $x\sqrt y + \sqrt y$ by $\sqrt x\sqrt{xy} + \sqrt{y}\cdot 1$ to make the reasoning more understandable. $\endgroup$ – amsmath Sep 2 '18 at 6:27
  • $\begingroup$ @amsmath I wrote that I used Cauchy-Schwarz. It's enough. $\endgroup$ – Michael Rozenberg Sep 2 '18 at 6:32
  • $\begingroup$ @ami_ba You are welcome! $\endgroup$ – Michael Rozenberg Sep 2 '18 at 6:33
  • $\begingroup$ @MichaelRozenberg Well, if you say so, then it must be right. $\endgroup$ – amsmath Sep 2 '18 at 12:57
2
$\begingroup$

After eliminating denominators and simplifying, the inequality reduces to:

$$ x^3y+xy^3-x^2y^2-2xy+1 \ge 0 \;\;\iff\;\; xy(x^2+y^2) - x^2y^2-2xy+1 \ge 0 $$

Using that $\,xy \ge 0\,$ and $\,\color{blue}{x^2+y^2 \ge 2xy}\,$, the above follows from:

$$ xy(x^2+y^2) - x^2y^2-2xy+1 \color{blue}{\ge 2x^2y^2}-x^2y^2-2xy+1 = (xy-1)^2 \ge 0 $$

$\endgroup$
  • $\begingroup$ See my first solution. I made it already. $\endgroup$ – Michael Rozenberg Sep 2 '18 at 6:56
  • $\begingroup$ @MichaelRozenberg Both our answers ultimately start from $\,x^3y+xy^3-x^2y^2-2xy+1 \ge 0\,$ as spelled out above, which follows from trivial algebraic manipulations. If you are insinuating that I somehow copied or duplicated your answer then, sorry, but I certainly did not, and hope no one (else) misconstrues it as such. $\endgroup$ – dxiv Sep 2 '18 at 7:11
  • 1
    $\begingroup$ No, I don't think so. Everything is fine! $\endgroup$ – Michael Rozenberg Sep 2 '18 at 8:38

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.