Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality $$\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$$?

share|cite|improve this question
Hint: Observe that it's sufficient to prove the inequality $\frac{\sqrt{x}}{1-x}-\frac{3\sqrt{3}}{2}x \geq 0$, when $x\in [0,1)$. Can you prove this one ? – Radu Titiu Aug 10 '12 at 7:12
@Radu Titu Why don't you post it as answer? – Norbert Aug 10 '12 at 8:53
@RaduTitiu thank you.the constraint $a+b+c=1$ can be weaken by $0<a,b,c<1$ – tan9p Aug 10 '12 at 9:37
and if I want to proof $$\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}+\frac{\sqrt{d}}‌​{1-d} \geq \frac{8}{3}$$for $a+b+c+d=1$?what's more,$$\sum_{i=1}^n\frac{\sqrt{a_i}}{1-a_i} \geq \frac{n^{\frac{3}{2}}}{n-1}$$for $\sum_{i=1}^n a_i=1$ – tan9p Aug 10 '12 at 9:49
@Norbert It was tagged as homework, so I thought that a hint would be more appropriate. the condition $a+b+c=1$ is essential in the proof; $0<a,b,c<1$ is not enough. – Radu Titiu Aug 10 '12 at 11:21

Although the function $f(x)=\sqrt{x}/(1-x)$ is not convex on (0,1), its tangent at $x=1/3$ lower bounds the function and passes through the origin. That is, for $0\leq x\leq 1$, we have $${\sqrt{x}\over 1-x}\geq {3\sqrt{3}\over2}\, x.$$

Plugging in $a,b,c$ and adding gives
$${\sqrt{a}\over 1-a}+{\sqrt{b}\over 1-b}+{\sqrt{c}\over 1-c}\geq {3\sqrt{3}\over2}.$$

Added reference: Exercise 8.1 on page 131 of The Cauchy-Schwarz Master Class by J. Michael Steele asks you to prove that for $p\geq 1$, and positive $a,b,c$, $${a^p\over b+c}+{b^p\over a+c}+{c^p\over a+b}\geq {1\over 2}\,3^{2-p}\,(a+b+c)^{p-1}.\tag1$$ He notes that for $p=1$ this reduces to Nesbitt's inequality. The inequality (1) fails for $0<p<p_c$, where $p_c={3\log(2)-2\log(3)\over \log(2)-\log(3)}=.29048$ by looking at $a=b=1/2$ and $c$ close to zero. But it holds again for $p=0$ by Jensen's inequality .

share|cite|improve this answer

This is as far as I got...

$\frac{(1-b)\sqrt{a}}{(1-b)(1-a)}$ + $\frac{(1-a)\sqrt{b}}{(1-a)(1-b)}$ + $\frac{\sqrt{c}}{(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $


$\frac{(1-b)\sqrt{a} + (1-a)\sqrt{b}}{(1-b)(1-a)}$ + $\frac{\sqrt{c}}{(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $


$\frac{(1-c)(1-b)\sqrt{a}+(1-c)(1-a)\sqrt{b}+(1-b)(1-a)\sqrt{c}}{(1-a)(1-b)(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $


$\frac{\sqrt{a}(1-b-c+bc)+\sqrt{b}(1-a-c+ac)+\sqrt{c}(1-a-b+ab)}{(c+ab)((1-c)} $ $\geq$ $\frac{3\sqrt{3}}{2} $


$\frac{\sqrt{a}(bc+a)+\sqrt{b}(ac+b)+\sqrt{c}(ab+c)}{(ab+c)(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $

Can anyone explain the rest to me? Can you say that (bc+a),(ac+b),(ab+c) $\leq$ 1 ? Also how would you formally say that, I mean I know that two fractions times each other is less than 1, but I don't know how to formally state that. Thanks!

share|cite|improve this answer

$\sqrt{a} = x, b=y^2, c=z^2 => x^2+y^2+z^2=1$ We have to prove $$\frac{x}{y^{2}+z^{2}}+\frac{y}{x^{2}+z^{2}}+\frac{z}{x^{2}+y^{2}}\geq \frac{3\sqrt{3}}{2}$$: $$\frac{2\sqrt{3}}{3}x\left ( y^{2}+z^{2} \right )\leq \left ( x^{2}+\frac{1}{3} \right )\left ( y^{2}+z^{2} \right )\leq \frac{\left ( x^{2}+y^{2}+z^{2}+\frac{1}{3} \right )^{2}}{4}=\frac{4}{9}$$ Do it the same for $\frac{2\sqrt{3}}{3}y\left ( z^{2}+x^{2} \right ), \frac{2\sqrt{3}}{3}z\left ( y^{2}+x^{2} \right )$ So: $$\frac{leftside}{\frac{2\sqrt{3}}{3}}=\sum \frac{x^{2}}{\frac{2\sqrt{3}}{3}x\left ( y^{2}+z^{2} \right )}\geq \frac{\sum x^{2}}{\frac{4}{9}}=\frac{9}{4}$$ $$\Rightarrow leftside=\sum \frac{x}{y^{2}+z^{2}}\geq \frac{3\sqrt{3}}{2}$$

share|cite|improve this answer
I asked my friend, he proved in different way: He proved $$\frac{x}{1-x^{2}}\geq \frac{3\sqrt{3}}{2}x^{2}$$ (i don't understand much :P) $$\frac{x}{1-x^2}=\sqrt{\frac{x^2}{(1-x^2)^2}}=\sqrt{\frac{2x^4}{2x^2(1-x^2)^2}}‌​$$ AM-GM: $$2x^2(1-x^2)^2 \le \left(\frac{2x^2+1-x^2+1-x^2}{3} \right)^3=\frac{8}{27}$$ => $$\frac{x}{1-x^2} \ge \sqrt{\frac{2x^4}{\frac{8}{27}}}=\frac{3\sqrt{3}}{2}x^2$$ – Xeing Nov 9 '12 at 6:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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