$a$, $b$, $c$ 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.
Best 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: $$\sum_{cyc}\sqrt{\frac{2}{a+b}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{2}{a+b}}\right)^2(a+b+c)}{a+b+c}}=$$ $$=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{1}{a+b}}\right)^2\sum\limits_{cyc}(a+b)}{a+b+c}}\geq\sqrt{\frac{(1+1+1)^3}{a+b+c}}=3\sqrt{\frac{3}{a+b+c}}.$$