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Wednesday, May 9, 2018

Inequality a,b,c

For a,b,c non-negative real numbers such that a+b+c=1, prove that: (3a+1)(3b+1)(3c+1) \geq 3 \sqrt{3}(\sqrt{a} + \sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})

Solution

With Cauchy we have: (3a + 1)(1 + 3b) \geq (\sqrt{3a} + \sqrt{3b})^2 = 3(\sqrt{a} + \sqrt{b})^2 Multiplying all of the similar inequalities, we get the desired result. Equality happens if and only if 3a = 1/(3b), 3b = 1/(3c), 3c = 1/(3a), which means a=b=c = 1/3.

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