INEQUALITY'S BLOG

September 7, 2010

Problem 49

Filed under: the new math — KKKVVV @ 10:54 pm

(Van Khea): Let $a, b, c>0$. Prove that

$\displaystyle \frac{a}{\sqrt{(a+2b)(a+2c)}}+\frac{b}{\sqrt{(b+2a)(b+2c)}}+\frac{c}{\sqrt{(c+2a)(c+2b)}}\geq 1$

Proof

From Van Khea’s inequality for $a, b, c, x, y, z, m, n, p>0$ and for $\alpha\geq 1, \beta , \gamma\geq 0$ such that $\alpha -\beta -\gamma=1$ We have

$\displaystyle \frac{a^{\alpha}}{x^{\beta}m^{\gamma}}+\frac{b^{\alpha}}{y^{\beta}n^{\gamma}}+\frac{c^{\alpha}}{z^{\beta}p^{\gamma}}\geq \frac{(a+b+c)^{\alpha}}{(x+y+z)^{\beta}(m+n+p)^{\gamma}}$
Let $\displaystyle P=\frac{a}{\sqrt{(a+2b)(a+2c)}}+\frac{b}{\sqrt{(b+2a)(b+2c)}}+\frac{c}{\sqrt{(c+2a)(c+2b)}}$
Therefore the inequality above can write that:
$\displaystyle P=\frac{a^2}{\sqrt{(a^2+2ab)(a^2+2ca)}}+\frac{b^2}{\sqrt{(b^2+2ab)(b^2+2bc)}}+\frac{c^2}{\sqrt{(c^2+2ca)(c^2+2bc)}}$

$\displaystyle P\geq \frac{(a+b+c)^2}{\sqrt{(a^2+b^2+c^2+2(ab+bc+ca)}\sqrt{(a^2+b^2+c^2+2(ab+bc+ca)}}=1$

$\Longrightarrow P\geq 1$