Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression:
a) $$P=\frac{4a}{b+c-a} + \frac{9b}{c+a-b} + \frac{16c}{a+b-c}$$
b) $$P=\frac{a^3}{2a+bc} + \frac{b^3}{2b+ac} + \frac{c^3}{2c+ab}, a+b+c=2$$
Only use AM-GM and Cauchy-Schwarz inequalities.
The problem here is that a, b, c are the 3 sides of the triangle so it has the triangle inequality so I can't find the equal condition.
Edit: My attempt:
a) I tried multiplying each fraction with the numerator variable, and then use Cauchy-Schwarz Engel's form and then it will become $$P\geq\frac{(2a+3b+4c)^2}{2ab+2bc+2ac-a^2-b^2-c^2}$$Let the RHS be Q. After that I use AM-GM with $2ab,2bc,2ac$ and then it will become $$Q\geq\frac{(2a+3b+4c)^2}{a^2+b^2+c^2}$$And then I'm stuck.
b) I tried dividing each fraction with the numerator, and then use Cauchy-Schwarz and get $$P\geq\frac{4}{6+\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}}$$And I'm stuck too.

The first problem.
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and by AM-GM we obtain:
$$\frac{4a}{b+c-a} + \frac{9b}{c+a-b} + \frac{16c}{a+b-c}=\frac{4(y+z)}{2x}+\frac{9(x+z)}{2y}+\frac{8(x+y)}{z}=$$
$$=\frac{2y}{x}+\frac{9x}{2y}+\frac{2z}{x}+\frac{8x}{z}+\frac{9z}{2y}+\frac{8y}{z}\geq6+8+12=26.$$
The equality occurs for $2y=3z,$ $z=2x$ and $3z=4y,$ id est, occurs, which says that we got a minimal value.
The second problem.
By C-S Schur and Muirhead we obtain: $$\sum_{cyc}\frac{a^3}{2a+bc}=\sum_{cyc}\frac{a^4}{2a^2+abc}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2+abc)}=$$
$$=\frac{2(a^2+b^2+c^2)^2}{(a+b+c)^2(a^2+b^2+c^2)+3abc(a+b+c)}=$$
$$=\frac{\sum\limits_{cyc}(2a^4+4a^2b^2)}{\sum\limits_{cyc}(a^4+2a^2b^2+2a^3b+2a^3c+5a^2bc)}=$$
$$=\frac{1}{2}+\frac{\sum\limits_{cyc}(3a^4-2a^3b-2a^3c+6a^2b^2-5a^2bc)}{2\sum\limits_{cyc}(a^4+2a^2b^2+2a^3b+2a^3c+5a^2bc)}=$$
$$=\frac{1}{2}+\frac{\sum\limits_{cyc}(3(a^4-a^3b-a^3c+a^2bc)+a^3b+a^3c+6a^2b^2-8a^2bc)}{2\sum\limits_{cyc}(a^4+2a^2b^2+2a^3b+2a^3c+5a^2bc)}\geq\frac{1}{2}.$$
The equality occurs for $a=b=c,$ which says that we got a minimal value.
