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 $\le 3/8.$

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1$

but you know from AM-GM $\frac{a^2+b^2}{2}\ge ab$

Note that,

$a^2+b^2+c^2=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{c^2+a^2}{2}\ge ab+bc+ca$

$\Rightarrow a^2+b^2+c^2+2(ab+bc+ca)\ge 3(ab+bc+ca)$

$\implies ab+bc+ca\le \frac13$

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

$ab+bc+ca \leq \frac{1}{3}$