For positive reals $a,\,b,\,c$, prove that $\displaystyle \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\gt 2$.

Hello readers!

In my previous blog post, I asked the readers to spot the factual mistake(s) that I might have or might not have made in the solution (of mine) to one delicious inequality problem.

Today, I am going to discuss with you the mistake that I intentionally made.

From the following steps that I made:

$\displaystyle \frac{a}{\sqrt{a}\sqrt{b+c}}+\frac{b}{\sqrt{b}\sqrt{c+a}}+\frac{c}{\sqrt{c}\sqrt{a+b}}$

$\displaystyle \gt \frac{a}{\frac{a+b+c}{2}}+\frac{b}{\frac{a+b+c}{2}}+\frac{c}{\frac{a+b+c}{2}}$ (from the AM-GM inequality)

I used the AM-GM inequalities below to get to the next step:

$\displaystyle \frac{a+(b+c)}{2}\ge \sqrt{a}\sqrt{b+c}$, with equality when $a=b+c$,

$\displaystyle \frac{b+(c+a)}{2}\ge \sqrt{b}\sqrt{c+a}$, with equality when $b=c+a$,

$\displaystyle \frac{c+(a+b)}{2}\ge \sqrt{c}\sqrt{a+b}$, with equality when $c=a+b$.

$\displaystyle \frac{a}{\frac{a+b+c}{2}}+\frac{b}{\frac{a+b+c}{2}}+\frac{c}{\frac{a+b+c}{2}}$

$\displaystyle \gt 2\left(\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\right)$

$\displaystyle = 2\left(\frac{a+b+c}{a+b+c}\right)$

$\displaystyle = 2$(Q.E.D.)

But to attain the equality, the conditions $a=b+c$, $b=c+a$, and $c=a+b$ must be satisfied. In other words, we got to have $a+b+c=0$. But we're told all $a,\,b$ and $c$ are positive reals, so their sum can't equal to zero.

This means we must omit the equality sign and thus, we will end up with the strict inequality:

 $\displaystyle \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\gt 2$

i.e. for positive reals $a,\,b,\,c$, the minimum for $\displaystyle \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$ isn't a $2$.