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

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The inequality $$\frac{a-\sqrt{bc}}{a+2b+2c}+\frac{b-\sqrt{ca}}{b+2c+2a}+\frac{c-\sqrt{ab}}{c+2a+2b}\ge 0$$ is proposed for proof with positive real numbers $a$, $b$, and $c$. The discussion emphasizes the need to demonstrate that each term in the sum is non-negative. Participants engage in exploring various approaches, including algebraic manipulations and potential applications of known inequalities. The goal is to establish the validity of the inequality under the given conditions. The thread highlights the collaborative effort to solve this mathematical challenge.
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Prove $$\frac{a-\sqrt{bc}}{a+2b+2c}+\frac{b-\sqrt{ca}}{b+2c+2a}+\frac{c-\sqrt{ab}}{c+2a+2b}\ge 0$$ holds for all positive real $a,\,b$ and $c$.
 
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anemone said:
Prove $$\frac{a-\sqrt{bc}}{a+2b+2c}+\frac{b-\sqrt{ca}}{b+2c+2a}+\frac{c-\sqrt{ab}}{c+2a+2b}\ge 0$$ holds for all positive real $a,\,b$ and $c$.
my solution:
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 $a=b=c$
 
Last edited:
Albert said:
my solution:
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:

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

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

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