Minimum value of $S=\frac{3a}{b+c}+\frac{4b}{a+c}+\frac{5c}{a+b}$ Find Minimum value of Minimum value of $$S=\frac{3a}{b+c}+\frac{4b}{a+c}+\frac{5c}{a+b}$$
My Try:
we have $$S=3 \times \left (\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)+\frac{b}{a+c}+\frac{2c}{a+b}$$
By Standard $AM$ $GM$ inequality we know that
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \ge \frac{3}{2}$$
Is it possible to proceed here?
 A: For any positive numbers $\alpha, \beta, \gamma$, we have
$$\frac{\alpha^2 a}{b+c} + \frac{\beta^2 b}{c+a} + \frac{\gamma^2 c}{a+b}
= (a+b+c)\left(\frac{\alpha^2}{b+c} + \frac{\beta^2}{c+a} + \frac{\gamma^2}{a+b}\right) - \left(\alpha^2+\beta^2+\gamma^2\right)$$
By Engel's form of Cauchy Schwarz, we have
$$\frac{\alpha^2}{b+c} + \frac{\beta^2}{c+a} + \frac{\gamma^2}{a+b}
\ge \frac{(\alpha+\beta+\gamma)^2}{(b+c)+(c+a)+(c+a)}
= \frac{(\alpha+\beta+\gamma)^2}{2(a+b+c)}
$$
This leads to
$$\frac{\alpha^2 a}{b+c} + \frac{\beta^2 b}{c+a} + \frac{\gamma^2 c}{a+b}
\ge \frac{(\alpha+\beta+\gamma)^2}{2} - \left(\alpha^2 + \beta^2 + \gamma^2\right)$$
and the equality holds when and only when
$$\alpha : \beta : \gamma = b+c : c+a : a+b$$
Subsitute $\alpha,\beta,\gamma$ by $\sqrt{3},\sqrt{4}$ and $\sqrt{5}$
We obtain
$$\frac{3 a}{b+c} + \frac{4 b}{c+a} + \frac{5 c}{a+b}
\ge \frac{(\sqrt{3} + \sqrt{4} + \sqrt{5})^2}{2} - 12
= \sqrt{15} + \sqrt{20} + \sqrt{12} - 6$$
and the equality is achievable when
$$a : b : c = -\sqrt{3}+\sqrt{4}+\sqrt{5} : \sqrt{3} - \sqrt{4} + \sqrt{5} : \sqrt{3}+\sqrt{4} - \sqrt{5}$$
This means the minimum value of $S$ is $\sqrt{15} + \sqrt{20} + \sqrt{12} - 6$.
