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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 $(a + b + c)$?

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I suggest you to write in LaTex – Riccardo.Alestra Feb 23 '12 at 12:15
I am a new user . I dont know how to write in latex.Any help will be welcome. – vikiiii Feb 23 '12 at 12:18
I could be wrong but here is my opinion: since $a,b,c$ have no special weights, I believe that there would be symmetry across them causing $a=b=c$. This when solved will result in $a+b+c=1$. Again, this has no mathematical foundation (other than intuition) so it could be wrong. – Inquest Feb 23 '12 at 12:36
@Nunoxic: $(a+b+c)^2=(a^2+b^2+c^2)+2(ab+bc+ca)\geq3(a^2b^2c^2)^{1/3}+2(ab+bc+ca)\geq1$. – Y.Z Feb 23 '12 at 12:44
up vote 6 down vote accepted

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 $a=b=c=\frac{1}{3}$.

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Thanks Daniel. You are awesum. – vikiiii Feb 23 '12 at 13:01

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