If $ 3a+2b+c=7$ then find minimum value of $ a^2+b^2+c^2$ Question:- If $ 3a+2b+c=7$ then find the minimum value of $ a^2+b^2+c^2 $.
I used vectors to solve this problem.
Let $$α=3\hat{i}+2\hat{j}+\hat{k}$$
$$β=a\hat{i}+b\hat{j}+c\hat{k}$$
Using Cauchy-Schwarz inequality
we have, $|α.β|\le |α| |β|$
$=|3a+2b+c|\le\sqrt{14}\sqrt{a^2+b^2+c^2}$
$= 7\le\sqrt{14}\sqrt{a^2+b^2+c^2}$
So, $a^2+b^2+c^2\ge \frac72$
Therefore, the minimum value of $a^2+b^2+c^2$ is $\frac72$
I want to know are there any other method to find the minimum value of
$a^2+b^2+c^2$ such as using inequalities and calculus by assuming function $f(x,y,z)=x^2+y^2+z^2$.
 A: There is a variant of the Cauchy-Schwarz inequality and its name is by a certain community of problem solvers as Titu’s lemma:
$a^2+b^2+c^2 = \dfrac{(3a)^2}{9}+\dfrac{(2b)^2}{4}+\dfrac{c^2}{1} \ge \dfrac{(3a+2b+c)^2}{9+4+1} = \dfrac{49}{14} = \dfrac{7}{2}$.
A: Also, $$a^2+b^2+c^2=\frac{1}{14}(3^2+2^2+1^2)(a^2+b^2+c^2)\geq\frac{1}{14}(3a+2b+c)^2=3.5.$$
The equality occurs for $$(3,2,1)||(a,b,c),$$ which says that we got a minimal value.
A: Think geometrically: The equation $3x+2y+z=7$ describes a plane in $\mathbb{R}^3$ with normal vector proportional to $(3,2,1)$. The closest point to the origin is therefore of the form $(a,b,c)=(3r,2r,r)$. From $3a+2b+c=7$ we have $9r+4r+r=7$, so $r=1/2$, and thus $a^2+b^2+c^2=(9+4+1)/4=7/2$.
(Rohan Nuckchady posted this approach as a comment beneath the OP. I either missed it, or was composing my answer when it appeared.)
A: Let $P(a, b, c)$ be any point on $(\pi): 3 x+2 y+z=7$.$\\\\$
$\\ OP^{2}=a^{2}+b^{2}+c^{2}$ is the minimum iff OP $\perp(\pi) \Leftrightarrow  \displaystyle O P =\left|\frac{3(0)+2(0)+7}{\sqrt{3^{2}+2^{2}+1^{2}}}\right| =\frac{7}{\sqrt{14}} =\sqrt{\frac{7}{2}}$
$\therefore$ the minimun value of $\displaystyle a^{2}+b^{2}+c^{2} \textrm{ is }\frac{7}{2}.$
