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.

Noting that
$$20 a^{2}+20 b^{2}+2 c^{2}-8(a b+b c+c a)=(2 a-2 b)^{2}+(4 b-c)^{2}+(4 a-c)^{2} \geqslant 0,
$$
we have
$$
\begin{aligned}
\quad 2\left(10 a^{2}+10 b^{2}+c^{2}\right)\geqslant 8 \quad \Rightarrow \quad 10 a^{2}+10 b^{2}+c^{2} \geqslant 4
\end{aligned}
$$
The minimum value of $10 a^{2}+10 b^{2}+c^{2}$ is $4$, which is attained if and only if$$
\left\{\begin{array}{l}
a=b \\
4 b=c \\
c=4 a
\end{array}\right.
 \Leftrightarrow 
a=b=\frac{1}{3} \text { and } c=\frac{4}{3}
$$
