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.
 A: Hint: If we additionally make the (reasonable, though not necessarily true) assumption that the minimum happens when $ a = b$, then reduce it to a 1 variable inequality and hence show that the minimum is 4 obtained at $ ( 1/3, 1/3, 4/3)$.
Note that we've not proven this is the minimum yet.
Use this to figure out what AM-GM's to create.
Modifying your approach, we have the following inequalities:
$a^2 + b^2 \geq 2ab \longrightarrow$ This reinforces $a=b$ at the equality case.
$Xa^2 + 0.5c^2 \geq 2\sqrt{X/2}ac \longrightarrow$ What does the (assumed) equality case tell us about $X$?
$Yb^2 + 0.5c^2 \geq 2\sqrt{Y/2}bc \longrightarrow$ What does the (assumed) equality case tell us about $Y$?
From the equality case, we require

 $ X = 8$, $ Y = 8$.

Then, weight the inequalities to get

 $ 2a^2 + 2b^2 \geq 4ab $
$ 8 a^2 + 0.5 c^2 \geq 4 ac $
$ 8b^2 + 0.5c^2 \geq 4bc $
 Summing them up gives:

$$ 10a^2 + 10b^2 + c^2 \geq 4(ab+bc+ca) = 4,$$
with equality when $ a = b = c/4, ab+bc+ca = 1$, IE $ ( 1/3, 1/3, 4/3)$.
(Thankfully, our initial assumption is true.)

I wanted to show you how to derive these inequalities (assuming that your stated approach could work). It isn't just "magic" or "by observation" or "by luck".
For the actual solution, you just need to write out the 3 AM-GM inequalities, sum them up, and verify the equality case.
Using this approach, try your hand at minimizing (say) $3a^2 + 2b^2 + c^2 $ given $ab+bc+ ca = 1$.
(Note: I've not actually done this, so I can't guarantee that the values will look nice.)
A: 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, $$\min_{ab+ac+bc=1}(10a^2+10b^2+c^2)=4.$$
A: 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$$
\left\{\begin{array}{l}
a=b \\
4 b=c \\
c=4 a
\end{array}\right.
 \Leftrightarrow 
a=b=\frac{1}{3} \text { and } c=\frac{4}{3}
$$
A: 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.
