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Question

Prove that abc3+bca3+cab3>1a+1b+1c, where a,b,c are different positive real numbers.

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Solution

Using A.M and G.M inequalities

abc3+bca3>2bac(1)

bca3+cab3>2cab(2)

cab3+abc3>2abc(3)

Adding all three equations, we get

abc3+bca3+cab3>bac+cab+abc(4)

Similarly, bac+cab>1a

cab+abc>1b

abc+bac>1c

From above equations, we get

bac+abc+cab>1a+1b+1c(5)

From (4) and (5),

abc3+bca3+cab3>1a+1b+1c



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