$\frac1{(a+1)^2}+ \frac1{(b+1)^2}\geq\frac 1{1+ab}$ for positive reals $a$ and $b$

Question

$\displaystyle\frac{1}{(a+1)^2}+ \displaystyle\frac{1}{(b+1)^2}\geq \displaystyle\frac{1}{1+ab}$ for positive reals $a$ and $b$

My idea was to apply Titu’s lemma but it didn’t work. Can you provide a proof that doesn’t simply use “brute force” a.k.a. SOS method? Thank you in advance

Answer 1

I don't know what the SOS method is, and therefore perhaps that this doesn't answer your question, but you could use the fact that$$\frac1{(a+1)^2}+\frac1{(b+1)^2}-\frac1{1+ab}=\frac{ab(a-b)^2+(1-ab)^2}{(a+1)^2(b+1)^2(1+ab)}.$$

Answer 2

By C-S $$\frac{1}{(a+1)^2}+\frac{1}{(b+1)^2}=\frac{b^2}{b^2(a+1)^2}+\frac{a^2}{a^2(b+1)^2}\geq$$ $$\geq\frac{(a+b)^2}{2a^2b^2+2ab(a+b)+a^2+b^2}=\frac{(a+b)^2}{(1+ab)(a+b)^2-ab((a-1)^2+(b-1)^2)}\geq\frac{1}{1+ab}.$$