For $a,b>0$ and $c\ge0$ prove that $\frac{2(a+b)}{3a+6b+9c}+\frac{6(b+c)}{5a+2b+3c}+\frac{3(a+c)}{2a+8b+6c}\ge\frac{3}{2}$

Question

For $a,b>0$ and $c\ge0$ prove that $\frac{2(a+b)}{3a+6b+9c}+\frac{6(b+c)}{5a+2b+3c}+\frac{3(a+c)}{2a+8b+6c}\ge\frac{3}{2}$

This question is easily solved if you say: $x=3a+6b+9c$, $y=5a+2b+3c$, $z=2a+8b+6c$.

From this we have that:

$2(a+b)=\frac{-x+y+z}{2}$

$6(b+c)=\frac{x-y+z}{2}$

$3(c+a)=\frac{x+y-z}{2}$

Hence it is enough to prove that $\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\ge6$ which holds true.

I couldn't solve this question, as I couldn't think of the substitution. Could you please explain to me intuitively why I should have thought of doing the particular $x,y,z$ substitutions?

Answer 1

It is not too the intuitive for me. The more intuitive substituion for me is

$$ \begin{align} x&=2(a+b)\\ y&=6(b+c)\\ z&=3(a+c) \end{align} $$

which will give us

$$ \frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\geq\frac{3}{2} $$

which is a well known inequality