How find the minimum of $ab+\frac{1}{a^2}+\frac{1}{b^2}$

Question

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

Answer 1

Here care must be taken while using AM-GM so that the constraint and equality condition is not violated. Hence rewrite the objective as:

$$\left(\frac12 ab+ \frac12 ab+\frac{1}{32a^2}+\frac{1}{32b^2}\right)+ \frac{31}{32}\left( \frac1{a^2}+\frac1{b^2}\right)$$ The first part is a sum of four terms with constant product. Hence this achieves minimum when each term is the same. Hence for this part, $$\implies 16ab^3=16a^3b=1 \implies a=b=\tfrac12$$

So for the first part, the minimum is $\frac12$. For the second part, we have $$\frac1{a^2}+\frac1{b^2} = \left(\frac1a-\frac1b\right)^2+\frac2{a^2b^2} \ge 0+ \frac8{(a+b)^2} \ge 8$$ which is also achieved when $a=b=\frac12$.

Hence the minimum of the objective within the constraints is achieved when $a=b=\frac12$, and the value is $\frac{33}4$.

Answer 2

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 $2\sqrt 2$

Answer 3

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 $a=b=\frac12$ $$S\ge\frac{33}4$$