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

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

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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 .

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This is as far as I got...

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

$\Leftrightarrow$

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

$\Leftrightarrow$

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

$\Leftrightarrow$

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

$\Leftrightarrow$

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

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!

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$\sqrt{a} = x, b=y^2, c=z^2 => x^2+y^2+z^2=1$ 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: $$\frac{leftside}{\frac{2\sqrt{3}}{3}}=\sum \frac{x^{2}}{\frac{2\sqrt{3}}{3}x\left ( y^{2}+z^{2} \right )}\geq \frac{\sum x^{2}}{\frac{4}{9}}=\frac{9}{4}$$ $$\Rightarrow leftside=\sum \frac{x}{y^{2}+z^{2}}\geq \frac{3\sqrt{3}}{2}$$

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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