<s>[INST] Using AM-GM Inequalities: For a,b,c all positive and $a+b+c=1$, prove that $ab+bc+ca\le 1/3$ Using AM-GM Inequalities.

For a,b,c all positive and $a+b+c=1$, prove that $ab+bc+ca\le 1/3$.

My attempt:
I took AM-GM of $a+b$ and $c$, and then $a$ and $b+c$ and so on.
Using that i got $\le 3/8.$
 [/INST] Since $\frac{a^2}{2}+\frac{b^2}{2}\geq 2 \sqrt{\frac{a^2}{2}\times\frac{b^2}{2}}=ab$
$1=(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=$
$=(\frac{a^2}{2}+\frac{b^2}{2})+(\frac{b^2}{2}+\frac{c^2}{2})+(\frac{c^2}{2}+\frac{a^2}
{2})+2ab+2bc+2ca\geq $
$\geq ab+bc+ca+2ab+2bc+2ca$
Therefore
$ab+bc+ca \leq \frac{1}{3}$
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