If $abc=1$ so $\sum\limits_{cyc}\frac{a}{\sqrt{a+b^2}}\geq\frac{3}{\sqrt2}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}}\geq\frac{3}{\sqrt2}$$
After substitution $a=\frac{y}{x}$... I tried C-S, but without success.
 A: I proved this inequality!!!
Let $a=\frac{x}{z}$, $b=\frac{y}{x}$ and $c=\frac{z}{y}$, where $x$, $y$ and $z$ are positives.
Hence, we need to prove that $\sum\limits_{cyc}\frac{x^2}{\sqrt{z(x^3+y^2z)}}\geq\frac{3}{\sqrt2}$.
Now by AM-GM $\sum\limits_{cyc}\frac{x^2}{\sqrt{z(x^3+y^2z)}}=\sum\limits_{cyc}\frac{2\sqrt2x^3}{2\sqrt{2x^2z(x^3+y^2z)}}\geq\sum\limits_{cyc}\frac{2\sqrt2x^3}{x^3+y^2z+2zx^2}$.
Thus, it remains to prove that $\sum\limits_{cyc}\frac{x^3}{x^3+y^2z+2zx^2}\geq\frac{3}{4}$, which is true, but my proof is still very ugly:
Let $x=\min{x,y,z\}$, $y=x+u$ and $z=x+v$.
Thus, we need to prove that:
$$8(u^2-uv+v^2)x^7+(2u^3+45u^2v+5uv^2+2v^3)x^6+$$
$$+(3u^4+40u^3v+153u^2v^2-40uv^3+3v^4)x^5+$$
$$+(5u^5+37u^4v+146u^3v^2+114u^2v^3-47uv^4+5v^5)x^4+$$
$$+(3u^6+20u^5v+130u^3v^3+21u^2v^4-18uv^5+3v^6)x^3+$$
$$+uv(7u^5+30u^4v+82u^3v^2+38u^2v^3-12uv^4+2v^5)x^2+$$
$$+u^2v^2(6u^4+20u^3v+27u^2v^2-6uv^2+v^4)x+u^5v^3(2u+5v)\geq0,$$ which is smooth.
A: Using Hölder's inequality, we have in general:
$$
\left( \frac{a}{\sqrt{X}}+\frac{b}{\sqrt{Y}}+\frac{c}{\sqrt{Z}}\right)^2(aX+bY+cZ) \geq (a+b+c)^3
$$
Substitute $\sqrt{a+b^2}, \sqrt{b+c^2}, \sqrt{c+a^2}$ for $X, Y, Z$:
$$
\left( \frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}} \right)^2 \geq \frac{(a+b+c)^3}{a^2+b^2+c^2+ab^2+bc^2+ca^2}
$$
Only we have to do is show (the right side) $\geq \frac{9}{2}$.
$$
f(a,b,c) \equiv 2(a+b+c)^3-9(a^2+b^2+c^2+ab^2+bc^2+ca^2) \\
= 12+2(a^3+b^3+c^3)-9(a^2+b^2+c^2)+6(a^2b+b^2c+c^2a)-3(ab^2+bc^2+ca^2)
$$
Since this is a symmetric polynomial, the min should be at $a=b=c(=1)$. Though this is not the strict proof.
Here, I tried Muirhead's inequality, for example,
$$
a^2b+b^2c+c^2a \geq ab^2+bc^2+ca^2
$$
but without success.
Instead, by partial differential, we can say that at $a=b=c=1$ we have the global minimum $0$ (for $a,b,c>0$. but the same thing if $a \geq b \geq c \space and \space b,c<0$). The inequality is true.
I hope it's OK as a hint.
