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

I think this solution is more simple.
From $$2a+b+3c=20\Leftrightarrow b=20-2a-3c$$
Then $$A=a^2+4\left(20-2a-3c\right)^2+c^2$$
$$=17a^2+48ac-320a+37c^2-480c+1600$$
$$=\frac{1}{17}(17a+24c-160)^2+\frac{1}{17}\left(53\left(c-\frac{240}{53}\right)^2+\frac{27200}{53}\right )$$
$$\ge \frac{1}{17}\cdot \frac{27200}{53}=\frac{1600}{53}$$
The equality occurs when $(a;b;c)=\left(\frac{160}{53};\frac{20}{53};\frac{240}{53}\right)$
