<s>[INST] 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 
 [/INST] Given:
$$ab+\frac 1{a^2}+\frac 1{b^2}=ab+\frac 1{a^2}+\frac 1{b^2}+\frac {2}{ab}-\frac {2}{ab}$$
$$\Rightarrow ab+\frac 1{a^2}+\frac 1{b^2}=ab+(\frac 1a-\frac 1b)^2+\frac 2{ab}$$
Let us analyse the RHS:
The minumum value of $(\frac 1a-\frac 1b)^2$ is 0 when a=b and the minumum value of $ab+\frac 2{ab}$ is $2\sqrt 2$(A.M.-G.M inequality). Hence the minimum value of the required expression is $2\sqrt 2$
</s>