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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,

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

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

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    $\begingroup$ from where do you got this? can you write down the whole sum please? $\endgroup$ Commented Aug 15, 2017 at 12:01
  • $\begingroup$ @Dr. Sonnhard Graubner, I got this problem from my friend, it's in C-S exercise. $\endgroup$
    – user403160
    Commented Aug 15, 2017 at 12:10
  • $\begingroup$ can you send me a link please? $\endgroup$ Commented Aug 15, 2017 at 12:43
  • $\begingroup$ @Dr. Sonnhard Graubner, It's paper work. Hint and source of the problem are not given. $\endgroup$
    – user403160
    Commented Aug 15, 2017 at 13:10

2 Answers 2

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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.

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  • $\begingroup$ Thank you for your help, Matsuda Toyomu. $\endgroup$
    – user403160
    Commented Aug 15, 2017 at 14:30
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I think your start is good.

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!

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    $\begingroup$ Thank you for your help, Michael Rozenberg. $\endgroup$
    – user403160
    Commented Aug 15, 2017 at 14:32
  • $\begingroup$ @carat You are welcome! $\endgroup$ Commented Aug 15, 2017 at 15:11

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