How find the minimum of $ab+\frac{1}{a^2}+\frac{1}{b^2}$ Let $a,b>0$ such that $$a+b\le 1$$ Find the minimum of
$$ab+\dfrac{1}{a^2}+\dfrac{1}{b^2}$$
My try: 
I can find this minimum，use Holder inequality
$$(\dfrac{1}{a^2}+\dfrac{1}{b^2})(a+b)^2\ge (1+1)^3=8$$
But 
$$ab\le \dfrac{(a+b)^2}{4}\le\dfrac{1}{4}$$
so for 
$$ab+\dfrac{1}{a^2}+\dfrac{1}{b^2}$$ minimum I can't find it,thank you 

By AM-GM
$$16ab+16ab+a^{-2}+b^{-2}\ge16$$
But
$$2\sqrt{ab}\le a+b\le 1\implies ab\le\frac14\implies -31ab\ge-\frac{31}4$$
Summing we get, with equality iff $a=b=\frac12$
$$S\ge\frac{33}4$$
