5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $$\dfrac{x^p}{y^q}+\dfrac{y^p}{z^q}+\dfrac{z^p}{x^q}\ge 3$$

$\endgroup$
3
$\begingroup$

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 $\sqrt{c}$.

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

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ But the inequality $\sum\limits_{cyc}\sqrt{a^4b^3c}\geq\sum\limits_{cyc}a^2bc$ is wrong. Try $a\rightarrow+\infty$ and $b<c$. $\endgroup$ – Michael Rozenberg Feb 14 at 6:14

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.