0
$\begingroup$

I want to minimize the following cyclic sum, where $a,b,c>0$: $$\sum_{\text{cyc}} \sqrt{\frac{a}{b+c}}=\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}.$$

More precisely, I think that

$$\sum_{\text{cyc}}\sqrt{\frac{a}{b+c}}\geq \frac{3}{\sqrt 2}$$

with equality if and only if $a=b=c$.


My work. I was able to prove $$\sum_{\text{cyc}}\sqrt{\frac{a}{b+c}}>2.$$ However, we have $\frac{3}{\sqrt 2}\approx2.1>2$.

Here is my proof of this weaker bound:

Note that for all $x,y,z>0$, $$(y+z-x)^2\geq 0,$$ so $$(x+y+z)^2\geq 4x(y+z),$$ and hence by taking the reciprocal and multiplying with $4x^2$, $$\frac{x}{y+z}\geq \left(\frac{2x}{x+y+z}\right)^2,$$ from which it follows that $$\sqrt{\frac{x}{y+z}}\geq\frac{2x}{x+y+z}$$ with equality if and only if $x=y+z$.

Hence we have (since equality can't occur for all three terms) $$\sum_{\text{cyc}} \sqrt{\frac{a}{b+c}}>\sum_{\text{cyc}} \frac{2a}{a+b+c}=2.$$

$\endgroup$
0
$\begingroup$

It's wrong. Try $c\rightarrow0^+$ and $a=b=1.$

| cite | improve this answer | |
$\endgroup$