If $abc=1$ so $\sum\limits_{cyc}\frac{a}{\sqrt{a+b^2}}\geq\frac{3}{\sqrt2}$

Question

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.

Answer 1

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.

Answer 2

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.