<s>[INST] 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.
 [/INST] For the second problem, notice that $ 2a+bc = (a+b+c)a+bc = (a+b)(a+c)$. This allows us to normalize the inequality and (mostly) forget the condition.   WTS
$$ \sum \frac{ a^3(b+c) } { (a+b)(b+c)(c+a) } \geq \frac{1}{2} $$
Approach 1: It is almost immediately obvious that by AM-GM / Muirhead (since there are no $a^4$ terms on the RHS, so $(3,1,0)$ majorizes everything),
$$ \sum a^3 (b+c) \geq \frac{1}{4} ( a+b)(b+c)(c+a)(a+b+c) $$
Hence the result follows.

Approach 2 via CS: WTS $$ \sum \frac{ a^3}{(a+b)(a+c) } \geq \frac{a+b+c } { 4 } $$
CS gives us
$$\sum \frac{a^4}{ a(a+b)(a+c) }  \geq \frac{ (a^2+b^2+c^2)^2 } { \sum a(a+b)(b+c) }  $$
Hence, it RTS (after cross multiplying and simplifying)
$$ \sum_{sym} 3/2 a^4 + 3 a^2b^2 \geq \sum_{sym}  2a^3b + 5/2 a^2bc$$
This is true because $(4, 0, 0 ) + (2, 2, 0 ) \geq 2 (3, 1, 0), \frac{1}{2}(4, 0, 0 )\geq \frac{1}{2} (2, 1, 1), (2, 2, 0 ) \geq (2, 1, 1)$
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