<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] $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1$
but you know from AM-GM $\frac{a^2+b^2}{2}\ge ab$
Note that,
$a^2+b^2+c^2=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{c^2+a^2}{2}\ge ab+bc+ca$
$\Rightarrow a^2+b^2+c^2+2(ab+bc+ca)\ge 3(ab+bc+ca)$
$\implies ab+bc+ca\le \frac13$
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