# 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 $x^2+y^2=1$

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

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.
• I think, we can't because $-\frac{12}{5+2xy}$ is negative. – Michael Rozenberg Apr 17 at 7:49