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$a$, $b$ and $c$ are real positive numbers satisfying

  • $\dfrac{1}{3} \le ab + bc + ca \le 1$ and
  • $abc \ge \dfrac{1}{27}$

What is the minimum possible value of $(a + b + c)$?

  • 0
    @Nunoxic: $(a+b+c)^2=(a^2+b^2+c^2)+2(ab+bc+ca)\geq3(a^2b^2c^2)^{1/3}+2(ab+bc+ca)\geq1$.2012-02-23

1 Answers 1

6

By the AM-GM inequality, $\sqrt[3]{abc}\leq \frac{a+b+c}{3}$.

Since $abc \geq \frac{1}{27}$, this implies that $\sqrt[3]{abc}\geq \frac{1}{3}$.

So $\frac{1}{3} \leq \frac{a+b+c}{3}$ and hence $1 \leq a+b+c$.

This minimum value is achieved in the symmetric case where $a=b=c=\frac{1}{3}$.

  • 0
    Thanks Daniel. You are awesum.2012-02-23