2016/117) If a,b,c are 3 positive numbers show that $\frac{a^2+1}{b+c} + \frac{b^2+1}{c+a} \frac{c^2+1}{a+b} >= 3$

we have $a^2+1 >= 2a$ hence $\frac{a^2+1}{b+c} >= 2\frac{a}{b+c}$
also $b^2+1 >= 2b$ hence $\frac{b^2+1}{c+a} >= 2\frac{b}{b+c}$
and $c^2+1 >= 2c$ hence $\frac{c^2+1}{a+b} >= 2\frac{c}{a+b}$
adding above 3 we get $\frac{a^2+1}{b+c} + \frac{b^2+1}{c+a} \frac{c^2+1}{a+b} >= 2(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b})$
now
If we can show that $2(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b})>=3$
we are through
we have $\frac{a}{b+c} + \frac{b}{c+a}$
= $\frac{ca + a^2 + b^2 + bc}{c+a}$
$>=\frac{ca + 2ab + b^2 + bc}{c+a}$ since $a^2+b^2>=2ab$
$= \frac{a(b+c) + b(c+a)}{(c+a)}$
Thus  $\frac{a}{b+c} + \frac{b}{c+a} >= \frac{a}{a+c} + frac{b}{b+c}$
Similarly, it can be proved that
$\frac{b}{c+a} + \frac{c}{a+b} >= \frac{b}{a+b} + frac{c}{c+a}$
and $\frac{c}{a+b} + \frac{a}{b+c} >= \frac{c}{b+c} + frac{a}{a+b}$
Adding corresponding sides of the above three inequalities, we get
$2(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b})>=3$