# Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$

Let $a,b,c \in (0, \infty)$, with $a+b+c=3$. How can I prove that: $$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3 ?$$.

I try to use Cauchy-Schwarz rewriting the inequality like :

$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\sum_{cyc}{\sqrt[4]{a^2b}})^2}{a+b+c}$$ but I don't obtain anything.

let $a=x^2,b=y^2,c=z^2$ $$\Longleftrightarrow x^2+y^2+z^2=3\Longrightarrow \dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge 3$$ note $$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2z^2x^2+2y^2z^2$$ use AM-GM inequality,then we have $$4\dfrac{x^2}{y}+2x^2y^2+x^4\ge (2\cdot 2+2+1)x^{\frac{2\cdot 2\times 2+2\cdot 2+2\cdot 2}{2\cdot 2+2+1}}=7x^{\frac{16}{7}}$$ and the same $$4\dfrac{y^2}{z}+2y^2z^2+y^4\ge 7y^{\frac{16}{7}}$$ $$4\dfrac{z^2}{x}+2z^2x^2+z^4\ge 7z^{\frac{16}{7}}$$ so $$4\sum\dfrac{x^2}{y}+(x^2+y^2+z^2)^2\ge 7\sum x^{\frac{16}{7}}$$ $$\Longleftrightarrow x^{\frac{16}{7}}+y^{\frac{16}{7}}+z^{\frac{16}{7}}\ge x^2+y^2+z^2$$

use AM-GM inequality we have $$7x^{\frac{16}{7}}+1=x^{\frac{16}{7}}+x^{\frac{16}{7}}+\cdots+x^{\frac{16}{7}}+1\ge 8\sqrt[8]{x^{\frac{16}{7}\cdot 7}}=8x^2$$ $$7\sum x^{\frac{16}{7}}+\sum x^2\ge \sum 8x^2$$ so $$\sum x^{\frac{16}{7}}\ge \sum x^2$$

In general,we have

$x^n+y^n+z^n=3,2p+q>2n,p,q,n\in N^{+}$,then $$\dfrac{x^p}{y^q}+\dfrac{y^p}{z^q}+\dfrac{z^p}{x^q}\ge 3$$

Rewrite it as $$\sum_{cyc} \frac{a}{\sqrt{b}} + (\sqrt{b} - 2\sqrt{a}) = \sum_{cyc} \frac{a}{\sqrt{b}} -\sum_{cyc} \sqrt{a} \ge 3 - \sum_{cyc} \sqrt{a} = \sqrt{3(a+b+c)} - \sum_{cyc} \sqrt{a}$$ LHS becomes $$\sum_{cyc} \frac{(\sqrt{a} - \sqrt{b})^2}{\sqrt{b}}$$ RHS becomes $$\frac{3(a+b+c) - (\sum_{cyc} \sqrt{a})^2}{\sqrt{3(a+b+c)} + \sum_{cyc} \sqrt{a}} = \frac{\sum_{cyc} (\sqrt{a} - \sqrt{b})^2}{\sqrt{3(a+b+c)} + \sum_{cyc} \sqrt{a}}$$ The inequality then follows since $\sqrt{3(a+b+c)} + \sum_{cyc} \sqrt{a} \ge \sqrt{a}, \sqrt{b}$ and $\sqrt{c}$.

• Hello,I think your last wrong。 – math110 Oct 24 '13 at 8:20
• @math110, where? – user27126 Oct 24 '13 at 8:21
• you last $$\Longleftrightarrow \sum_{cyc}\dfrac{(\sqrt{a}-\sqrt{b})^2}{\sqrt{b}}\ge \dfrac{\sum_{cyc}(\sqrt{a}-\sqrt{b})^2}{\sqrt{3(a+b+c)}+\sum_{cyc}\sqrt{a}}$$ – math110 Oct 24 '13 at 8:25
• @math110, so what is the mistake here? I don't see it. – user27126 Oct 24 '13 at 8:26
• It's not equivalent $$\sqrt{3(a+b+c)}+\sum_{cyc}\sqrt{a}\ge\sqrt{b}$$ and $\sqrt{a}$ and $\sqrt{c}$ – math110 Oct 24 '13 at 8:26

Since $\left(1/\sqrt{x}\right)^{\prime\prime} = \frac{3}{4 x^{5/2}} > 0$ for $x > 0$, we have that $f(x)=\frac{1}{\sqrt{x}}$ is convex.

Then, by Jensen's inequality we have:

$$\tfrac{1}{a+b+c}\times \left( \frac{a}{\sqrt{b}} + \frac{b}{\sqrt{c}} + \frac{c}{\sqrt{a}} \right) \geq \left( \frac{ab+bc+ca}{a+b+c} \right)^{-1/2}$$

Here $9=(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \geq 3\times (ab+bc+ca)$ $(*)$, so we actually get $3\geq ab+bc+ca$ and therefore:

$$\tfrac{1}{a+b+c}\times \left( \frac{a}{\sqrt{b}} + \frac{b}{\sqrt{c}} + \frac{c}{\sqrt{a}} \right) \geq \left( \frac{ab+bc+ca}{a+b+c} \right)^{-1/2} \geq \left( \frac{3}{a+b+c} \right)^{-1/2} = 1$$

Since we have $a+b+c=3$, we are done.

$(*)$ This follows from $2\times \{a^2+b^2+c^2-(ab+bc+ca)\} = (a-b)^2 +(b-c)^2 + (c-a)^2 \geq 0$

By Holder’s inequality (with $$\sum$$ denoting cyclic sums): $$\left(\sum \frac{a}{\sqrt b}\right)^2\left(\sum ab \right)\geqslant (a+b+c)^3=27$$ So it is enough to show $$\sum ab \leqslant 3=\frac13(a+b+c)^2$$, which is well known.

Assume that $x,y,z\in\mathbb{R}^+$ and $x^2+y^2+z^2=3$. We want to prove: $$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq 3.$$ Since $f(w)=\frac{1}{w}$ is a convex function on $\mathbb{R}^+$, we have: $$\frac{x^2}{3}f(y)+\frac{y^2}{3}f(z)+\frac{z^2}{3}f(x)\geq f\left(\frac{x^2y+y^2 z+z^2 x}{3}\right),$$ so: $$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq \frac{9}{x^2y+y^2 z+z^2 x},$$ and it is sufficient to prove $x^2y+y^2z+z^2x\leq 3$. In virtue of the Cauchy-Schwarz inequality, $$x^2 y + y^2 z + z^2 y \leq \sqrt{(x^2+y^2+z^2)(x^2y^2+y^2z^2+z^2y^2)},$$ so it is sufficient to prove $x^2y^2+y^2z^2+z^2y^2\leq 3$, that is equivalent to $x^4+y^4+z^4\geq 3$, that is trivial in virtue of the Cauchy-Schwarz inequality, again.

We need to prove that $$\sum_{cyc}\frac{a^2}{b}\geq3,$$ where $$a$$, $$b$$ and $$c$$ are positives such that $$a^2+b^2+c^2=3.$$

Indeed, $$\sum_{cyc}\frac{a^2}{b}-3=\sum_{cyc}\left(\frac{a^2}{b}-2a+b\right)-\left(3-a-b-c\right)=$$ $$=\sum_{cyc}\frac{(a-b)^2}{b}-\frac{9-(a+b+c)^2}{3+a+b+c}=\sum_{cyc}\frac{(a-b)^2}{b}-\frac{3(a^2+b^2+c^2)-(a+b+c)^2}{3+a+b+c}=$$ $$\sum_{cyc}\frac{(a-b)^2}{b}-\frac{\sum\limits_{cyc}(a-b)^2}{3+a+b+c}=\sum_{cyc}\frac{(a-b)^2(3+a+c)}{b(3+a+b+c)}\geq0$$ and we are done!

With homogenizazion: We need to prove for all $$a,b,c>0$$: $$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq\sqrt{3}\sqrt{a+b+c}.$$ By squaring this is equivalent to $$\frac{\sum_{cyc} a^3c+2\sum_{cyc} a^2 b^\frac32 c^\frac12}{abc}\geq \frac{3\sum_{cyc} a^2bc}{abc}$$ which follows immediately from AM-GM, for example for the first sum, use the cyclical versions of $$2ab^3+4a^3c+bc^3\geq 7a^2bc$$ by AM-GM.

• But the inequality $\sum\limits_{cyc}\sqrt{a^4b^3c}\geq\sum\limits_{cyc}a^2bc$ is wrong. Try $a\rightarrow+\infty$ and $b<c$. – Michael Rozenberg Feb 14 at 6:14