$a$ 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.