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.


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.


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.

  • $\begingroup$ The inequality $2(a+b+c)^3\geq9\sum\limits_{cyc}(a^2+ab^2)$ is wrong. Try $a=4$, $b=\frac{1}{5}$ and $c=\frac{5}{4}$. $\endgroup$ Apr 2 '17 at 3:14
  • $\begingroup$ confirmed. I was wrong... $\endgroup$
    – user426180
    Apr 4 '17 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.