# Inequality : $\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}\geq \frac{3}{5}$

Let $a$, $b$ and $c$ be positive real numbers. Prove that: $$\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}+\frac{\sqrt{b^3a}}{2\sqrt{c^3b}+3ca}+\frac{\sqrt{c^3b}}{2\sqrt{a^3c}+3ab}\geq \frac{3}{5}$$

My attempt :

Substitute $a=x^2, b=y^2, c=z^2$, we have to prove that

$\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2} \geq \frac{3}{5}$

By C-S,

$\left(\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2}\right)\left(\displaystyle\sum_{cyc}(2y^2x+3y^2z^2)\frac{x}{z}\right) \geq \left(\displaystyle\sum_{cyc}x^2\right)^2$

$\left(\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2}\right)\left(\displaystyle\sum_{cyc}(x^2y^2+3y^2xz\right) \geq \left(\displaystyle\sum_{cyc}x^2\right)^2$

• from where do you got this? can you write down the whole sum please? Commented Aug 15, 2017 at 12:01
• @Dr. Sonnhard Graubner, I got this problem from my friend, it's in C-S exercise. Commented Aug 15, 2017 at 12:10
• can you send me a link please? Commented Aug 15, 2017 at 12:43
• @Dr. Sonnhard Graubner, It's paper work. Hint and source of the problem are not given. Commented Aug 15, 2017 at 13:10

Let $a = x^2, b = y^2, c=z^2$, where $x, y, z > 0$. By C-S, we have $$\sum_{cyc} \frac{x^3 z}{2 y^3 x + 3y^2z^2} \sum_{cyc} \frac{2yx + 3z^2}{zx} \geq \left( \sum_{cyc} \frac{x}{y} \right)^2.$$ But $$\sum_{cyc} \frac{2yx + 3z^2}{zx} = 5 \sum_{cyc} \frac{x}{y}.$$ Combining them we get $$\sum_{cyc} \frac{x^3 z}{2 y^3 x + 3y^2z^2} \geq \frac{1}{5} \sum_{cyc} \frac{x}{y}.$$ It remains to apply AM-GM.
Let $a=x^2$, $b=y^2$ and $c=z^2$, where $x$, $y$ and $z$ are positives.
Hence, by C-S $$\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}=\sum_{cyc}\frac{x^3z}{y^2(3z^2+2xy)}=\frac{1}{x^2y^2z^2}\sum_{cyc}\frac{x^5z^3}{3z^2+2xy}=$$ $$=\frac{1}{x^2y^2z^2}\sum_{cyc}\frac{x^6z^4}{xz(3z^2+2xy)}\geq\frac{(x^3z^2+y^3x^2+z^3y^2)^2}{x^2y^2z^2\sum\limits_{cyc}(3x^3y+2x^2yz)}.$$ Thus, it's enough to prove that $$5(x^3z^2+y^3x^2+z^3y^2)^2\geq3\sum_{cyc}3x^5y^3z^2+2x^4y^3z^3)$$ or $$\sum_{cyc}(5x^6z^4+x^5y^3z^2-6x^4y^3z^3)\geq0,$$ which is true by Rearrangement: $$\sum_{cyc}x^6z^4=x^4y^4z^4\sum_{cyc}\frac{x^2}{y^4}\geq x^4y^4z^4\sum_{cyc}\frac{x^2}{x^4}=$$ $$=x^4y^4z^4\sum_{cyc}\frac{1}{x^2}\geq x^4y^4z^4\sum_{cyc}\frac{1}{xy}=\sum_{cyc}x^4y^3z^3$$ and $$\sum_{cyc}x^5y^3z^2=x^3y^3z^3\sum_{cyc}\frac{x^2}{z}\geq x^3y^3z^3\sum_{cyc}\frac{x^2}{x}=\sum_{cyc}x^4y^3z^3.$$ Done!