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a+b+c=1,prove \; that\sum \frac{\sqrt{a}}{1-a}\geqslant \frac{3\sqrt{3}}{2};prove:\left ( a+b+c \right )^{3}=\left ( \sum \sqrt[3]{\frac{\sqrt{a}}{1-a}} \ast \sqrt[3]{\frac{\sqrt{a}}{1-a}} \ast \sqrt[3]{{a^{2}(1-a)^{2}}} \right )^{3}
Hi，Thank you，I have proved p=1/2 the inequality is right.
I want to know p=1/3 the inequality is right too? how prove? pln3/2>ln9/8 the inequality is right too?right?
Thank you.
How do you do it?help me.
yes,thank you!
yes,thank you!
\frac{3^{(2-p)}}{2}
Hi;
\frac{3^(2-p)}{2}
Not
3^{\frac{(2-p)}{2}}
thank you!
\frac{a^p}{1-a}+\frac{b^p}{1-b}+\frac{c^p}{1-c}\geq \frac{3^(2-p)}{2}
Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
a^p/(1-a)+b^p/(1-b)+c^p/(1-c)≥3^(2-p)/2
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