<s>[INST] 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.
 [/INST] Let $k$ be the minimal value.
Thus, $k>0$ and the inequality $$10a^2+10b^2+c^2\geq k(ab+ac+bc)$$ or $$c^2-k(a+b)c+10a^2+10b^2-kab\geq0$$ is true for any reals $a$,$b$ and $c$ (because after homogenization the condition $ab+ac+bc=1$ is not relevant already), which says $$k^2(a+b)^2-4(10a^2+10b^2-kab)\leq0,$$ which gives
$$(40-k^2)a^2-2(2k+k^2)ab+(40-k^2)b^2\geq0,$$ for which we need $40-k^2>0$ and $$(2k+k^2)^2-(40-k^2)^2\leq0$$ or $$(k+20)(k+5)(k-4)\leq0$$ or
$$0<k\leq4,$$ which gives that a maximal value of $k$, for which the inequality $$10a^2+10b^2+c^2\geq k(ab+ac+bc)$$ is true it's $4$.
The equality occurs for $a=b$ and $c=\frac{k(a+b)}{2}=4a$.
Id est, $$\min_{ab+ac+bc=1}(10a^2+10b^2+c^2)=4.$$
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