Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression

Question

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.

Answer 1

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.

Answer 2

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.

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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 $(4, 0, 0 ) + (2, 2, 0 ) \geq 2 (3, 1, 0), \frac{1}{2}(4, 0, 0 )\geq \frac{1}{2} (2, 1, 1), (2, 2, 0 ) \geq (2, 1, 1)$