How prove $\left(1+\frac{4a}{b+c}\right)\left(1+\frac{4b}{c+a}\right)\left(1+\frac{4c}{a+b}\right)\ge 25$ let $a,b,c>0$ show that
$$\left(1+\dfrac{4a}{b+c}\right)\left(1+\dfrac{4b}{c+a}\right)\left(1+\dfrac{4c}{a+b}\right)\ge 25$$
It seem hard to prove AM-GM.Cauchy-Schwarz 

Hint
On full expansion, this reduces to a combination of Schur's inequality $\sum a^3 +3abc\geqslant \sum ab(a+b)$ and $abc>0$, so the inequality is in fact strict. 
