minimum of $a^2+4b^2+c^2$ given $2a+b+3c=20$ 
If $a,b,c\in\mathbb{R}$ and $2a+b+3c=20.$ Then minimum value of $a^2+4b^2+c^2$ is 

what i try
Cauchy schwarz inequality
$$(a^2+(2b)^2+c^2)(2^2+\frac{1}{2^2}+3^2)\geq (2a+b+3c)^2$$
How do i solve it without Cauchy schwarz inequality Help me please

Let $(a,b,c)=\left(\frac{160}{53},\frac{20}{53},\frac{240}{53}\right).$
Thus, we get a value $\frac{1600}{53}.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$a^2+4b^2+c^2\geq\frac{1600}{53}\cdot\left(\frac{2a+b+3c}{20}\right)^2$$ or
$$37a^2-16(b+3c)a+208b^2-24bc+17c^2\geq0,$$ for which it's enough to prove that
$$64(b+3c)^2-37(208b^2-24bc+17c^2)\leq0,$$ which is
$$(12b-c)^2\geq0.$$
Done!
