Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$

Question

$a$ 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.

Answer 1

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!

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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

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

Answer 6

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.