Solve $ \sum_{cyc}\frac{ab}{\sqrt{ab+bc}} \leq \frac{1}{\sqrt{2}}$

Question

Let $a, b, c$ be positive real numbers such that $a+b+c = 1$. Prove that $$ \displaystyle\sum_{cyc}\frac{ab}{\sqrt{ab+bc}} \leq \frac{1}{\sqrt{2}}$$

My attempted work :

By C-S, $$ (ab+ac)(1+1) \geq (\sqrt{ab}+\sqrt{bc})^2$$

$$\sqrt{2} \sqrt{ab+bc} \geq \sqrt{ab}+\sqrt{bc}$$

$$\frac{\sqrt{2} ab}{ \sqrt{ab}+\sqrt{bc}} \geq \frac{ab}{\sqrt{ab+bc}}$$

$$\frac{ab}{\sqrt{ab+bc}} \leq \frac{\sqrt{2} ab}{ \sqrt{ab}+\sqrt{bc}}$$

multiply through by $\sqrt{2}$

$$\displaystyle\sum_c \frac{\sqrt{2} ab}{\sqrt{ab+bc}} \leq \displaystyle\sum_c \frac{ 2ab}{ \sqrt{ab}+\sqrt{bc}} = \displaystyle\sum_c \frac{ ab}{ \sqrt{ab}+\sqrt{bc}} + \displaystyle\sum_c \frac{ bc}{ \sqrt{ab}+\sqrt{bc}} = \displaystyle\sum_c \frac{ ab+bc}{ \sqrt{ab}+\sqrt{bc}}$$

Please suggest, how to show that

$$\displaystyle\sum_c \frac{ ab+bc}{ \sqrt{ab}+\sqrt{bc}} \leq \frac{1}{\sqrt{2}}\sqrt{2} = 1 = a+b+c $$

Can we just use basic inequalities ?

Answer 1

Your solution is wrong because $$ 2 \displaystyle\sum_{cyc} \frac{ ab}{ \sqrt{ab}+\sqrt{bc}} \neq \displaystyle\sum_{cyc} \frac{ ab}{ \sqrt{ab}+\sqrt{bc}} + \displaystyle\sum_{cyc}\frac{ bc}{ \sqrt{ab}+\sqrt{bc}} $$

My proof:

By C-S $$\left(\sum_{cyc}\sqrt{\frac{a^2b}{a+c}}\right)^2\leq(ab+ac+bc)\sum_{cyc}\frac{a}{a+c}=$$ $$=(ab+ac+bc)\left(3-\sum_{cyc}\frac{c}{a+c}\right)=(ab+ac+bc)\left(3-\sum_{cyc}\frac{c^2}{ac+c^2}\right)\leq$$ $$\leq(ab+ac+bc)\left(3-\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}\right).$$ Thus, it remains to prove that $$(ab+ac+bc)\left(3-\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}\right)\leq\frac{(a+b+c)^2}{2}.$$ Now, let $a^2+b^2+c^2=k(ab+ac+bc)$.

Thus, $k\geq1$ and we need to prove that $$3-\frac{k+2}{k+1}\leq\frac{k+2}{2}$$ or $$k(k-1)\geq0.$$ Done!