<s>[INST] 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
 [/INST] Put $b'=2b$ so that we minimize $a^2+(b')^2+c^2$ subject to $4a+b'+6c=40$. Now, since the point $x=(a,b',c)$ is on the plane $P: 4x+y+6z=40$, its distance from the origin is minimized when it is parallel to the normal vector $n=(4,1,6)$ to the plane $P$. If we solve $x=tn=(4t,t,6t)\in P$ for $t\in\Bbb R$, then we get $53t=40$ and $x=\frac{40}{53}(4,1,6)$ follows. Hence we obtain
$$
a^2+(b')^2+c^2 \ge |x|^2=\frac{1600}{53}
$$ is the minimum.
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