# Prove this inequality for given conditions

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 $$(a^2 + b^2) (ab-a-b +2) \ge a^2b^2$$

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: $$\sum_{cyc}\frac{1}{(x+1)^2}=\sum_{cyc}\frac{y}{(x\sqrt{y}+\sqrt{y})^2}\geq\sum_{cyc}\frac{y}{(xy+1)(x+y)}=\frac{1}{xy+1}.$$

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

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:

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

• See my first solution. I made it already. – Michael Rozenberg Sep 2 '18 at 6:56
• @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. – dxiv Sep 2 '18 at 7:11
• No, I don't think so. Everything is fine! – Michael Rozenberg Sep 2 '18 at 8:38