minimum of $a^2+4b^2+c^2$ given $2a+b+3c=20$

Question

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

Answer 1

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.

Answer 2

An alternative way of solving the problem is using Lagrange multipliers, other than that (i.e. without using the CS inequality), I don't see a simple solution, and neither do I see a reason to look for such a solution.

Answer 3

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!

Answer 4

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 $(a;b;c)=\left(\frac{160}{53};\frac{20}{53};\frac{240}{53}\right)$