<s>[INST] Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$ $a$ and $b$ are both positive Real Numbers .
I saw this in a math olympiad, and within two days i couldn't solve it.
There is a simple case where:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
We can solve it like this:
$$(a-1)^{2}\geq 0$$
$$\Leftrightarrow a^{2}-2a+1 \geq0$$
$$\Leftrightarrow a^{2}+1 \geq2a$$
$$\Leftrightarrow \frac{a^{2}+1}{a} \geq2$$
With the same method:
$$\frac{b^{2}+1}{b} \geq2 $$
And when we add the two inequalities:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
But in case where:
$$\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$$
I can’t solve it,
Thank you for your help.
 [/INST] You can use your method, making $(a-1)^2$ and $(b-1)^2$ appearing was a nice move.
$$\frac{a^2+1}{b}+\frac{b^2+1}{a}-4=\frac 1{\underbrace{ab}_{>0}}(\underbrace{a^3+a+b^3+b-4ab}_E)$$
Now use $2ab\le a^2+b^2$
$E\ge a^3+a+b^3+b-2a^2-2b^2=a(a-1)^2+b(b-1)^2\ge 0$
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