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

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 ?

• What does the notation $\sum_ {cyc}$ mean? I have noticed that it is a usual notation here in ME. I gave a google and found nothing about it. Commented Jul 16, 2017 at 14:45
• @MathOverview, The notation $\sum_{cyc} f(a, b, c)$ refers to the cyclic sum $f(a,b,c) + f(b,c,a) + f(c,a,b)$. Commented Jul 16, 2017 at 14:48

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!

• Regarding my work, I've edited the point you mentioned, is it correct now ? Commented Jul 16, 2017 at 15:01
• @carat It's still wrong, but by your first step I can end the proof. It's ugly, but it gives a proof. I think, it's better, see my proof. Commented Jul 16, 2017 at 15:06
• Your solution is elegant. Thank you very much ! Commented Jul 16, 2017 at 15:39
• @carat You are welcome! Commented Jul 16, 2017 at 16:23

Some years ago, I came up with a proof.

Proof: By using the Cauchy-Schwarz inequality, we have \begin{align} &\frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}}\\ \le \ & \sqrt{(xy+yz+zx)\Big(\frac{xy}{xy+yz}+\frac{yz}{yz+zx}+\frac{zx}{zx+xy}\Big)}\\ = \ & \sqrt{xy + \frac{x^2z}{x+z} + yz + \frac{xy^2}{y+x} + xz + \frac{yz^2}{z+y}}\\ \le \ & \sqrt{xy + \frac{x(x+z)^2/4}{x+z} + yz + \frac{y(y+x)^2/4}{y+x} + xz + \frac{z(z+y)^2/4}{z+y}}\\ = \ & \sqrt{\frac{1}{4}(x^2+y^2+z^2)+\frac{5}{4}(xy+yz+zx)}\\ \le \ & \sqrt{\frac{1}{2}(x+y+z)^2}\\ = \ & \frac{1}{\sqrt{2}}. \end{align}