If $ab+bc+ca=1$ what is the minimum value of $10a^2+10b^2+c^2$? 
$a,b,c$ are positive real numbers such that $ab+bc+ca=1$. What is the minimum value of $10a^2+10b^2+c^2$?

I want to solve this problem without using Lagrange multipliers or calculus. I tried the following with some basic inequalities:
From AM-GM inequality $$5a^2+5b^2\geq10ab\\ 5b^2+\frac 12c^2\geq\sqrt{10}bc\\ \frac 12c^2+5a^2\geq\sqrt{10}ca$$
Summing them gives $$10a^2+10b^2+c^2\geq\sqrt{10}(\sqrt{10}ab+bc+ca)$$
But this doesn't help.

Another way.
By C-S
$$10a^2+10b^2+c^2=$$
$$=\sqrt{\left(8a^2+\frac{c^2}{2}+8b^2+\frac{c^2}{2}+2a^2+2b^2\right)\left(\frac{c^2}{2}+8a^2+\frac{c^2}{2}+8b^2+2b^2+2a^2\right)}\geq\sqrt{(2ac+2ac+2bc+2bc+2ab+2ab)^2}=\sqrt{16}=4.$$
The equality occurs, when
$$\left(2\sqrt2a,\frac{c}{\sqrt2},2\sqrt2b,\frac{c}{\sqrt2},\sqrt2a,\sqrt2b\right)||\left(\frac{c}{\sqrt2},2\sqrt2a,\frac{c}{\sqrt2},2\sqrt2b,\sqrt2b,\sqrt2a\right),$$ which gives $a=b=\frac{c}{4}$, which says that we got a minimal value.
