# Math Help - Inequality

1. ## Inequality

Let $a,b,c$ be positive real numbers. Prove that
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\geq\d frac{3}{2}$

2. Originally Posted by math sucks
Let $a,b,c$ be positive real numbers. Prove that
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\geq\d frac{3}{2}$
This happens to be well-known.

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$

$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a +b} \geq \frac{9}{2}$.

$\Delta = (a+b+c)\left( \frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b} \right)\geq \frac{9}{2}$

But, $\frac{3}{\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b} } \leq \frac{2(a+b+c)}{3}$ by HM-AM.

Thus, $\Delta \geq (a+b+c)\cdot \frac{9}{2(a+b+c)} = \frac{9}{2}$.