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Let $a,b,c\in \Bbb R^+$ such that $a+b+c=abc$. Prove that $$\frac{a}{c\sqrt{a^2+1}}+\frac{b}{a\sqrt{b^2+1}}+\frac{c}{b\sqrt{c^2+1}}\ge \frac{3}{2}$$


Idea 1.From $a+b+c=abc\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1$. Let $\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\rightarrow \left(x;y;z\right)$

So i need to prove $\frac{z}{\sqrt{1+x^2}}+\frac{y}{\sqrt{z^2+1}}+\frac{x}{\sqrt{y^2+1}}\ge \frac{3}{2}$

By AM-GM $\frac{x}{\sqrt{y^2+1}}=\frac{x}{\sqrt{\left(x+y\right)\left(y+z\right)}}\ge \frac{2x}{x+2y+z}$

$$LHS\ge 2\sum _{cyc}\frac{x}{x+2y+z}=2\sum _{cyc}\frac{x^2}{x^2+2xy+xz}\ge 2\frac{\left(x+y+z\right)^2}{\sum _{cyc}x^2+\sum _{cyc}3xy}$$

Or $4\left(x+y+z\right)^2\ge 3\left(x^2+y^2+z^2+3xy+3yz+3xz\right)$

Or $x^2+y^2+z^2\ge xy+yz+xz$ (true)

Idea 2. By Holder $$\left(\sum _{cyc}\frac{a}{c\sqrt{a^2+1}}\right)\left(\sum _{cyc}\frac{a}{c\sqrt{a^2+1}}\right)\sum _{cyc}\left(c^2\left(a^2+1\right)a\right)\ge \left(\sum _{cyc}a\right)^3$$

I will prove the inequality $\frac{\left(a+b+c\right)^3}{c^2a\left(a^2+1\right)+a^2b\left(b^2+1\right)+b^2c\left(c^2+1\right)}\ge \frac{3}{2}$

Or $\frac{abc\left(a+b+c\right)^3}{\left(a+b+c\right)\left(a^2b^3+b^2c^3+c^2a^3\right)+abc\left(a^2b+b^2c+c^2a\right)}\ge \frac{3}{2}$

I tried $SOS$ but failed help me improve "idea 2" use Holder

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We need to prove that $$\sum_{cyc}\frac{a}{c\sqrt{a^2+\frac{abc}{a+b+c}}}\geq\frac{3}{2}$$ or $$\sum_{cyc}\sqrt{\frac{a(a+b+c)}{c^2(a+b)(a+c)}}\geq\frac{3}{2}$$ or $$\sum_{cyc}\sqrt{\frac{a^2b}{c(a+b)(a+c)}}\geq\frac{3}{2}$$ or $$\sum_{cyc }\sqrt{a^3b^2(b+c)}\geq\frac{3}{2}\sqrt{abc(a+b)(a+c)(b+c)}$$ or $$\sum_{cyc }\sqrt{(a^3b^3+a^3b^2c)}\geq\frac{3}{2}\sqrt{abc(a+b)(a+c)(b+c)}.$$ Now, let $ab=z$, $ac=y$ and $bc=x$.

Thus, we need to prove that $$\sum_{cyc}\sqrt{z^3+z^2x}\geq\frac{3}{2}\sqrt{(x+y)(x+z)(y+z)}.$$ By Holder $$\left(\sum_{cyc}\sqrt{z^3+z^2x}\right)^2\sum_{cyc}\frac{z}{z+x}\geq(x+y+z)^3.$$ Id est, it's enough to prove that $$4(x+y+z)^3\geq9\prod_{cyc}(x+y)\sum_{cyc}\frac{z}{z+x}$$ or $$4(x+y+z)^3\geq9\sum_{cyc}x(x+z)(y+z)$$ or $$4(x+y+z)^3\geq9\sum_{cyc}(2x^2y+x^2z+xyz).$$ Now, by Rearrangement easy to show that $$x^2y+y^2z+z^2x+xyz\leq\frac{4}{27}(x+y+z)^3.$$ Thus, it's enough to prove that $$4(x+y+z)^3\geq\frac{4}{3}(x+y+z)^3+9\sum_{cyc}\left(x^2y+x^2z+\frac{2}{3}xyz\right)$$ or $$\frac{8}{3}(x+y+z)^3\geq9(x+y)(x+z)(y+z)$$ or $$\left(\frac{x+y+x+z+y+z}{3}\right)^3\geq(x+y)(x+z)(y+z),$$ which is true by AM-GM.

Done!

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