# $a, b, c$ are three reals such that $ab + bc + ca = 3abc$. Prove that $\sum_{cyc}\sqrt{2(a + b)} \ge \sum_{cyc}\sqrt{\frac{a^2 + b^2}{a + b}} + 3$.

$$a$$, $$b$$, $$c$$ are three reals such that $$ab + bc + ca = 3abc (a + b, b + c, c + a > 0)$$. Prove that $$\large \sum_{cyc}\sqrt{2(a + b)} \ge \sum_{cyc}\sqrt{\frac{a^2 + b^2}{a + b}} + 3$$

Let $$ax = by = cz = 1 \implies x + y + z = 3$$

It is necessitated that it should be sufficient to prove that $$\sum_{cyc}\sqrt{\frac{2(x + y)}{xy}} \ge \sum_{cyc}\sqrt{\frac{x^2 + y^2}{xy(x + y)}} + 3$$

$$\implies \sum_{cyc}\left[\sqrt{\frac{2(x + y)}{xy}} - \sqrt{\frac{x^2 + y^2}{xy(x + y)}}\right] \ge x + y + z$$

$$\implies \sum_{cyc}\frac{\dfrac{x^2 + 4xy + y^2}{(x + y)\sqrt{xy}}}{\sqrt{2(x + y)} + \sqrt{\dfrac{x^2 + y^2}{x + y}}} \ge x + y + z$$

$$\implies \sum_{cyc}\left[\frac{\dfrac{x^2 + 4xy + y^2}{(x + y)\sqrt{xy}}}{\sqrt{2(x + y)} + \sqrt{\dfrac{x^2 + y^2}{x + y}}} - \frac{x + y}{2}\right] \ge 0$$

$$\implies \sum_{cyc}\frac{(x + y)\sqrt{xy(x^2 + y^2)(x + y)} + (x + y)^2\sqrt{2xy(x + y)} - 2(x^2 + 4xy + y^2)}{2(x + y)\sqrt{xy} \cdot \left[\sqrt{2(x + y)} + \sqrt{\dfrac{x^2 + y^2}{x + y}}\right]} \ge 0$$

The above attempt was written with the help of WolframAlpha, since it is humanly impossible to evaluate the above expression.

I would like a solution which can be done in an examinable setting, which could be done without any extra calculating from computers.

• Your attempt is wrong because in your problem $xy$ can be negative. I have a proof for positives $a$, $b$ and $c$. Dec 10, 2019 at 17:17
• Post it as an answer then. Honestly, I think that our teacher might have given the condition a little bit too far-fetched. Dec 11, 2019 at 0:27

## 1 Answer

It's wrong.

Try $$a=b=\frac{1}{4}$$ and $$b=-\frac{1}{5}.$$

The hint for positive variables.

Since by C-S $$\sqrt{2(a+b)^2}=\sqrt{(1+1)(a^2+b^2+2ab)}\geq\sqrt{a^2+b^2}+\sqrt{2ab},$$ it's enough to prove that $$\sum_{cyc}\sqrt{\frac{2ab}{a+b}}\geq3\sqrt{\frac{3abc}{ab+ac+bc}}.$$ Now, after replacing $$a$$ and $$\frac{1}{a},$$ $$b$$ on $$\frac{1}{b}$$ and $$c$$ on $$\frac{1}{c}$$ we need to prove that $$\sum_{cyc}\sqrt{\frac{2}{a+b}}\geq3\sqrt{\frac{3}{a+b+c}},$$ which is true by Holder: $$\sum_{cyc}\sqrt{\frac{2}{a+b}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{2}{a+b}}\right)^2(a+b+c)}{a+b+c}}=$$ $$=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{1}{a+b}}\right)^2\sum\limits_{cyc}(a+b)}{a+b+c}}\geq\sqrt{\frac{(1+1+1)^3}{a+b+c}}=3\sqrt{\frac{3}{a+b+c}}.$$

• How to do the last step? We need to prove $$\sum_{cyc} \sqrt{\frac{2ab}{a+b}}\geq 3$$ but it is not obvious to me because we still need to use the condition
– user729882
Dec 12, 2019 at 11:09
• @user729882 I added something. See now. Dec 12, 2019 at 11:21