# 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.

I would start with noticing that $$a^{2} + 1 \geq 2\sqrt{a^{2}} = 2a$$.

Similarly, we have that $$b^{2} + 1 \geq 2b$$.

Then we have that \begin{align*} \frac{a^{2} + 1}{b} + \frac{b^{2} + 1}{a} \geq 2\left(\frac{a}{b} + \frac{b}{a}\right) \geq 4\sqrt{\frac{a}{b}\times\frac{b}{a}} = 4 \end{align*}

Hopefully this helps!

• How did you figure out the formula in the middle
– PNT
Commented Jan 23, 2021 at 21:57
• It is a simple rearrangement. Indeed, we have that $$\frac{a^{2}+1}{b} + \frac{b^{2}+1}{a} \geq \frac{2a}{b} + \frac{2b}{a} = 2\left(\frac{a}{b} + \frac{b}{a}\right)$$ The proposed solution is the repeated application of the AM-GM inequality. Is this what you are talking about? Commented Jan 23, 2021 at 21:59
• Yes thank you for your help
– PNT
Commented Jan 23, 2021 at 22:01
• You are welcome ! Commented Jan 23, 2021 at 22:01

One can actually just use AM-GM directly: $$\frac{a^2}b+\frac1b+\frac{b^2}a+\frac1a\geq 4\sqrt[4]{\frac{a^2}{b}\cdot\frac{1}{b}\cdot\frac{b^2}a\cdot\frac1a}=4\sqrt[4]{1}=4.$$

Hint: 1) Use AM-GM : $$x^2+ 1\ge 2x$$ 2) UseAM-GM : $$t+\dfrac{1}{t} \ge 2$$

I assume that $$a,b$$ are positive real numbers. Then we may assume $$a\geq b$$, hence $$a^2+1\geq b^2+1$$ and $$\frac{1}{a}\leq\frac{1}{b}$$, so the rearrangement inequality gives $$\frac{a^2+1}{b}+\frac{b^2+1}{a}\geq\frac{a^2+1}{a}+\frac{b^2+1}{b}\geq4$$ where the last inequality holds by what you already showed.

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 $$2ab\le a^2+b^2$$

$$E\ge a^3+a+b^3+b-2a^2-2b^2=a(a-1)^2+b(b-1)^2\ge 0$$

I would do this the following way--no knowledge of AM--GT needed: Assume WLOG $$a \ge b$$. Then this inequality is clearly true for $$a \ge 4$$, as $$\frac{a^2+1}{b} > a$$ for all $$a,b \in \mathbb{N}$$ satisfying $$a \ge b$$. So now we are left to check the cases where both $$a,b \le 3$$; $$a\ge b$$. This can be done directly quite easily.