Prove that, for $a+b+c+d+e=3$,
\[ \left( \frac{a}{1-a} \right) \left( \frac{b}{1-b} \right) \left( \frac{c}{1-c} \right) \left( \frac{d}{1-d} \right) \left( \frac{e}{1-e} \right) \ge \left( \frac{3}{2} \right)^{5} \]
a + b + c + d + e = 3
ধনঞ্জয় বিশ্বাস
Re: a + b + c + d + e = 3
It's false; try $a=2, b=c=d=e=1/4$.
It is also false when $a,b,c,d,e\in (0,1)$; try $a=.0009, b=c=d=e=(3-.0009)/4$.
It is also false when $a,b,c,d,e\in (0,1)$; try $a=.0009, b=c=d=e=(3-.0009)/4$.
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