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 $x^2+y^2=1$

Question

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.

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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

Answer 1

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.