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

Question

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

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

score 6 · Accepted Answer · answered Feb 23 '12 at 13:00

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

answered Feb 23 '12 at 13:00

Daniel Pietrobon

1 Answers1