About the fact that $\frac{a^2}{b} + \frac{b^2}{a} + 7(a + b) \ge 8\sqrt{2(a^2 + b^2)}$

There's a math competition I participated yesterday (19/3/2019).

In these kinds of competitions, there will always be at least one problem about inequalities.

Now this year's problem about inequality is very easy. I am more interested in last year's problem, which goes by the following:

If $$a$$ and $$b$$ are positives then prove that $$\frac{a^2}{b} + \frac{b^2}{a} + 7(a + b) \ge 8\sqrt{2(a^2 + b^2)}$$

Now I know what you are thinking. It's a simple problem.

By the Cauchy - Schwarz inequality, we have that $$\dfrac{a^2}{b} + \dfrac{b^2}{a} \ge \dfrac{(a + b)^2}{a + b} = a + b$$. $$\implies \dfrac{a^2}{b} + \dfrac{b^2}{a} + 7(a + b) \ge (a + b) + 7(a + b) = 8(a + b) \le 8\sqrt{2(a^2 + b^2)}$$

And you have fallen into the traps of the people who created the test. Almost everyone in last year's competition did too.

But someone came up with an elegant solution to the problem. He was also the one winning the contest.

You should have 15 minutes to solve the problem. That's what I also did yesterday.

• What do you want from this problem? – Word Shallow Mar 20 at 16:22

Squaring the given inequality and factorizing we get $${\frac { \left( {a}^{2}+18\,ba+{b}^{2} \right) \left( a-b \right) ^{4 }}{{b}^{2}{a}^{2}}} \geq 0$$ which is true.
\begin{align*} \dfrac{a^2}{b} + \dfrac{b^2}{a} + 7(a + b) &= \dfrac{a + b}{ab} \cdot (a^2 + b^2 - ab + 7ab)\\ &= \dfrac{a + b}{ab} \cdot [(a + b)^2 + 4ab]\\ &\ge \dfrac{a + b}{ab} \cdot 4(a + b)\sqrt{ab}\\ &= \dfrac{(a + b)^2}{2\sqrt{2ab \cdot (a^2 + b^2)}} \cdot 8\sqrt{2(a^2 + b^2)}\\ &\ge \dfrac{a^2 + b^2 + 2ab}{a^2 + b^2 + 2ab} \cdot 8\sqrt{2(a^2 + b^2)}\\ &= 8\sqrt{2(a^2 + b^2)} \end{align*}
Let $$a^2+b^2=2k^2ab,$$ where $$k>0$$.
Thus, by AM-GM $$k\geq1$$ and we need to prove that: $$\frac{\sqrt{(a+b)^2}(a^2-ab+b^2)}{ab}+7\sqrt{(a+b)^2}\geq8\sqrt{2(a^2+b^2)}$$ or $$\sqrt{2k^2+2}(2k^2-1)+7\sqrt{2k^2+2}\geq16k$$ or $$\sqrt{2k^2+2}(k^2+3)\geq8k,$$ which is true by AM-GM: $$\sqrt{2k^2+2}(k^2+3)\geq\sqrt{4k}\cdot4\sqrt{k^2\cdot1^3}=8k.$$ Done!