Inequality Challenge: Prove $\ge 0$ for All $a,b,c$

[anemone]

Prove \(\displaystyle \frac{a-\sqrt{bc}}{a+2b+2c}+\frac{b-\sqrt{ca}}{b+2c+2a}+\frac{c-\sqrt{ab}}{c+2a+2b}\ge 0\) holds for all positive real $a,\,b$ and $c$.

 

[Albert1]

Prove \(\displaystyle \frac{a-\sqrt{bc}}{a+2b+2c}+\frac{b-\sqrt{ca}}{b+2c+2a}+\frac{c-\sqrt{ab}}{c+2a+2b}\ge 0\) holds for all positive real $a,\,b$ and $c$.

my solution:

Spoiler

by using $AP\geq GP$ we have:

$\dfrac{a-\sqrt {bc}}{a+2b+2c}+\dfrac{b-\sqrt {ca}}{b+2c+2a}+\dfrac{c-\sqrt {ab}}{c+2a+2b}>
\\

\dfrac{a-{(b+c)/2}}{2a+2b+2c}+\dfrac{b-{(c+a)/2}}{2a+2b+2c}+\dfrac{c-{(a+b)/2}}{2a+2b+2c}=
0\\$
equality holds when $a=b=c$

 

[anemone]

my solution:

Spoiler

by using $AP\geq GP$ we have:

$\dfrac{a-\sqrt {bc}}{a+2b+2c}+\dfrac{b-\sqrt {ca}}{b+2c+2a}+\dfrac{c-\sqrt {ab}}{c+2a+2b}>
\\

\dfrac{a-{(b+c)/2}}{2a+2b+2c}+\dfrac{b-{(c+a)/2}}{2a+2b+2c}+\dfrac{c-{(a+b)/2}}{2a+2b+2c}=
0\\$
equality holds when $a=b=c$

Thanks Albert for participating!

Here is my solution:

Spoiler

\(\displaystyle \begin{align*}\frac{a-\sqrt{bc}}{a+2b+2c}+\frac{b-\sqrt{ca}}{b+2c+2a}+\frac{c-\sqrt{ab}}{c+2a+2b}&\ge \frac{a-\frac{b}{2}-\frac{c}{2}}{a+2b+2c}+\frac{b-\frac{c}{2}-\frac{a}{2}}{b+2c+2a}+\frac{c-\frac{b}{2}-\frac{a}{2}}{c+2a+2b}\\&\ge \frac{a-\frac{b}{2}-\frac{c}{2}}{3\sqrt{a^2+b^2+c^2}}+\frac{b-\frac{c}{2}-\frac{a}{2}}{3\sqrt{a^2+b^2+c^2}}+\frac{c-\frac{b}{2}-\frac{a}{2}}{3\sqrt{a^2+b^2+c^2}}\\&= \frac{1}{3\sqrt{a^2+b^2+c^2}}\left(a-\frac{b}{2}-\frac{c}{2}+b-\frac{c}{2}-\frac{a}{2}+c-\frac{b}{2}-\frac{a}{2}\right)\\&=\frac{1}{3\sqrt{a^2+b^2+c^2}}\left(1+b+c-a-b-c\right)\\&=0\,\,\,\,\text{Q.E.D.}\end{align*}\)

The first step follows from the AM-GM inequality that says \(\displaystyle \frac{b+c}{2}\ge \sqrt{bc}\).

The second step follows from the Cauchy-Schwarz inequality that tells \(\displaystyle a+2b+2c\le\sqrt{1+2^2+2^2}\sqrt{a^2+b^2+c^2}\)