Minimize it!

Given expression is :

$\frac{3a}{b+c} + 3$ + $\frac{4b}{c+a} + 4$ + $\frac{5c}{a+b} + 5 - 12$

$\frac{3a + 3b + 3c}{b+c}$ + $\frac{4b + 4c + 4a}{c+a}$ + $\frac{5c + 5a + 5b}{a+b} - 12$

$[\frac{3}{b+c} + \frac{4}{c + a} + \frac{5}{a+b}]$ $[a + b + c]$ $-12$

Now , using Titu's Lemma , we get ,

$[\frac{3}{b+c} + \frac{4}{c + a} + \frac{5}{a+b}][a + b + c] -12 \ge \frac {\left[\sqrt3 + \sqrt4 + \sqrt5 \right]^{2}}{2(a+b+c)}(a+ b+c) -12$

$[\frac{3}{b+c} + \frac{4}{c + a} + \frac{5}{a+b}][a + b + c] -12 \ge \frac {\left[\sqrt3 + \sqrt4 + \sqrt5 \right]^{2}}{2} -12$

$[\frac{3}{b+c} + \frac{4}{c + a} + \frac{5}{a+b}][a + b + c] -12 \ge 5.809$

$\text{Try my problem “A cubic inequality"}$

$\frac{3 a}{b+c}+\frac{5 c}{a+b}+\frac{4 b}{a+c}=(a+b+c) \left(\frac{5}{a+b}+\frac{4}{a+c}+\frac{3}{b+c}\right)-12=\frac{1}{2} \left(\frac{5}{x}+\frac{3}{y}+\frac{ 4}{z}\right) (x+y+z)-12$

where $x=a+b,y=b+c,z=a+c$ . Then by setting $X=\frac{x}{\sqrt{5}}$ , etc.

$\frac{1}{2} \left(\frac{\sqrt{5}}{X}+\frac{\sqrt {3}}{Y}+\frac{2}{Z}\right) \left(\sqrt{5} X+\sqrt{3} Y+2 Z\right)-12\geq \frac{1}{2} \left(2+\sqrt{3}+\sqrt{5}\right)^2 \left(\frac{1}{X}\right)^{\frac{\sqrt{5}}{2+\sqrt{3}+\sqrt{5}}} X^{\frac{\sqrt{5}}{2+\sqrt{3}+\sqrt{ 5}}} \left(\frac{1}{Y}\right)^{\frac{\sqrt{3}}{2+\sqrt{3}+\sqrt{5}}} Y^{\frac{\sqrt{3}}{2+\sqrt{3}+\sqrt{ 5}}} \left(\frac{1}{Z}\right)^{\frac{2}{2 +\sqrt{3}+\sqrt{5}}} Z^{\frac{2}{2+\sqrt{3}+\sqrt{5}}}-12 =\frac{1}{2} \left(2+\sqrt{3}+\sqrt{5}\right)^2-1 2$

where I used the weighted AM - GM.

Anybody have a simpler solution?