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Positive numbers a, b, c satisfy $1/a + 1/b + 1/c = 3$.

a) Prove that $abc \ge 1$.

b) Prove that $(a+b)(a+c)(b+c) \ge 8$. When does the equality hold?

I want to say that I will be using conjugates for part a to prove the inequality is true. I am not sure about part b...

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Hints:

a) Use $GM \ge HM$

b) Use $a+b \ge 2\sqrt{ab}$ for each factor.

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This is a classic example of AM-GM style inequality: so you have: ab + bc + ca = 3abc. And ab + bc + ca >= 3(abc)^(2/3). So 3abc >= 3(abc)^(2/3) and this implies (abc)^3 >= (abc)^2 so you have abc >= 1. And (a + b)(b + c)(c + a) >= 2(ab)^(1/2)*2(bc)^(1/2)*2(bc)^(1/2) = 8abc >= 8 since abc >= 1.

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