Question

If a,b, c are positive real numbers, prove that
$\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}\ge a+b+c$

Answer 1

Since $(b+c{)}^2+(b-c{)}^2=2(b^2+c^2)$,

it follows that $b^2+c^2\ge \frac{1}{2}(b+c{)}^2$, i.e., $\frac{b^2+c^2}{b+c}\ge \frac{1}{2}(b+c)$
Similarly $\frac{c^2+a^2}{c+a}\ge \frac{1}{2}(c+a)$
$\frac{a^2+b^2}{a+b}\ge \frac{1}{2}(a+b)$
Adding corresponding sides of the above inequalities, we have
$\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}\ge a+b+c$