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Let $a,b,c \gt 0$. Prove that (Using Cauchy-Schwarz) : $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3$$

I tried to use Cauchy-Schwarz in the following form $$\sqrt{Ax}+\sqrt{By}+\sqrt{Cz}\leq \sqrt{(A+B+C)(x+y+z)}\tag{1}.$$

I wrote $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}}=\frac{\sqrt{2a(c+a)(a+b)}+\sqrt{2b(b+c)(a+b)}+\sqrt{2c(b+c)(c+a)}}{\sqrt{(a+b)(b+c)(c+a)}}.$$

and then I applied on $(1)$: \begin{eqnarray} A &=& 2a(c+a) &\mbox{and}& x=a+b;\\ B &=& 2b(a+b) &\mbox{and}& y=b+c;\\ C &=& 2c(b+c) &\mbox{and}& z=c+a , \end{eqnarray}

but I did not obtain anything. Thanks for your help. :)

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  • $\begingroup$ Did you mean for a, b, c to be > 0 and also integers? $\endgroup$
    – Amzoti
    Commented Sep 9, 2012 at 0:28

4 Answers 4

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You got it wrong. It should be: $$ \sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}} \leq 3 $$

See my comment on Inequality. $\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3$ for a proof.

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Denote $$S = \sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{c+a}} + \sqrt{\frac{c}{a+b}}$$

The Inequality doesn't hold. Clearly, taking $b=c=0.05, a=5$ implies $$\sqrt{2}S>\sqrt{\frac{2a}{b+c}} = 10 > 3$$ By this method, it is easily shown that no upper bound exists.

For a lower bound, note firstly that when $a = b, c \rightarrow 0$, $S \rightarrow 2$. We'll prove that $S \ge 2$ for non negative reals $a,b,c$.

We have, $$\frac{a+b+c}{2a} = \frac 12 \left(\frac{b+c}{a} + 1 \right) \ge \sqrt{\frac{b+c}{a}} \\ \implies \sqrt{\frac{a}{b+c}} \ge \frac{2a}{a+b+c}$$ Adding up the other two similar inequalities, we get the result.

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  • $\begingroup$ For $a = b$ and $c\to 0$, $S\to 2\sqrt 2$ $\endgroup$
    – AlbertH
    Commented Sep 9, 2012 at 1:11
  • $\begingroup$ @AlbertH Thanks! Edited. $\endgroup$ Commented Sep 9, 2012 at 2:26
  • $\begingroup$ I ask you to give an idea or a solution using Cauchy-Schwarz. Thanks anyway-I observed you like using analysis to prove inequalities-but I need Cauchy-Schwarz inequlity. $\endgroup$
    – Iuli
    Commented Sep 9, 2012 at 6:59
  • $\begingroup$ @Iuli I'm pretty surprised to see you thinking that AM-GM is analysis.. Anyway, how about I say that $\left(\frac{b+c}{a} + 1 \right) \left(1 + \frac{b+c}{a} \right) \ge \left(2 \sqrt{\frac{b+c}{a}} \right)^2$ by Cauchy Schwarz, so all that I'm doing is applying Cauchy Schwarz here.. ;) $\endgroup$ Commented Sep 9, 2012 at 21:43
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We will begin with an observation. Consider the function $$f(a,b,c)=\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} .$$ It has the property that $$f(ta,tb,tc)=f(a,b,c).$$ We may thus assume without loss of generality that $$a+b+c=1$$, by setting $$t=\frac{1}{a+b+c}.$$ Under this assumption, we may rewrite the function as $$\sqrt{\frac{2a}{1-a}}+\sqrt{\frac{2b}{1-b}}+\sqrt{\frac{2c}{1-c}}$$ If $a=1-\epsilon$ for $\epsilon>0$, $b+c=\epsilon$. Now note that in order for these square roots to be defined over the reals, then $0\leq a,b,c<1$, which means that $b,c<\epsilon$. Now for such an $\epsilon$, $$\sqrt{\frac{2b}{1-b}}, \sqrt{\frac{2c}{1-c}}$$ are both close to zero, and $$\sqrt{\frac{2a}{1-a}}$$ is large.

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  • $\begingroup$ and an idea using Cauchy-Schwarz where can I find ? $\endgroup$
    – Iuli
    Commented Sep 9, 2012 at 6:58
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By AM-HM, $\frac{2(a+b+c)}{3}\ge\frac{3}{\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}$$\implies(2a+2b+2c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge3^2$.

By Cauchy-Schwarz, $(2a+2b+2c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\left(\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{a+c}}\right)^2$.

Thus, $\left(\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{a+c}}\right)^2\le3^2$$\implies\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{a+c}}\le3$.

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