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Question

If a,b,c are positive real numbers, then prove that [(1+a)(1+b)(1+c)]7>77a4b4c4.

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Solution

[(1+a)(1+b)(1+c)]7>77a4b4c4
i.e.,(1+a+ab+abc)>7(abc)4/7
Since, AM>GM
a+b+c+3(abc)1/3
ab+bc+ac3(a2b2c2)1/3
Hence,
1+a+ab+abc1+3(abc)1/3+3(abc)2/3+abc1+a+ab+abc1+(abc)1/3+(abc)1/3+(abc)1/3+(abc)2/3+(abc)2/3+(abc)2/3+abc
So,
LHS7[1(abc)1/3(abc)1/3(abc)1/3(abc)2/3(abc)2/3(abc)2/3(abc)]1/77[(abc)9/3(abc)]1/7
7(abc)4/7

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