$a;b;c\in \mathbb{R}^+$ such that $abc=1$. Prove : $P=\frac{ab}{2b+c}+\frac{bc}{2c+a}+\frac{ca}{2a+b}\geq 1$

$a;b;c\in \mathbb{R}^+$ such that $abc=1$. Prove : $P=\frac{ab}{2b+c}+\frac{bc}{2c+a}+\frac{ca}{2a+b}\geq 1$

Thanks :)

I have proved that : $P=\sum \frac{ab}{2b+c}\Rightarrow P=\sum \frac{1}{2bc+c^{2}}$

Use Cauchy-Schwarz : $\Rightarrow P=\sum \frac{1}{2bc+c^{2}}\geq \frac{(1+1+1)^{2}}{(a+b+c)^{2}}=\frac{9}{(a+b+c)^{2}}$

But : $a+b+c\geq 3\sqrt[3]{abc}=3\Rightarrow (a+b+c)^{2}\geq 9\Rightarrow \frac{9}{(a+b+c)^{2}}\leq 1$

!?

Solution 1:

let $$a=\dfrac{z}{y},b=\dfrac{x}{z},c=\dfrac{y}{x}$$

then $$\dfrac{ab}{2b+c}=\dfrac{x^2z}{2x^2y+y^2z}$$ so Use cauchy-Schwarz inequality $$\left(\dfrac{x^2z}{2x^2y+y^2z}+\dfrac{y^2x}{2y^2z+z^2x}+\dfrac{z^2y}{2z^2x+x^2y}\right)(z(2x^2y+y^2z)+x(2y^2z+z^2x^2)+y(2z^2x+x^2y))\ge (xz+yx+zy)^2$$ it suffices to prove that $$\dfrac{(xy+yz+xz)^2}{(z(2x^2y+y^2z)+x(2y^2z+z^2x^2)+y(2z^2x+x^2y))}\ge 1$$ $$\Longleftrightarrow (xy+yz+xz)^2\ge (z(2x^2y+y^2z)+x(2y^2z+z^2x^2)+y(2z^2x+x^2y))$$ since $$(z(2x^2y+y^2z)+x(2y^2z+z^2x^2)+y(2z^2x+x^2y))=2xyz(x+y+z)+(x^2y^2+y^2z^2+x^2z^2)$$ it's obvious is true.