If $ax+by+cz=k$,prove that the minimum value of $x^2+y^2+z^2$ is $\frac{k^2}{a^2+b^2+c^2}$

Question

If $ax+by+cz=k$,prove that the minimum value of $x^2+y^2+z^2$ is $\frac{k^2}{a^2+b^2+c^2}$.

I know that this problem can be solved by Cauchy Schwartz.How can i find its minimum value using multivariable calculus or by other methods.Please help me.

Answer 1

$$(x^2+y^2+z^2)(a^2+b^2+c^2)-(ax+by+cz)^2 = (ay-bx)^2+(bz-cy)^2+(cx-az)^2 \ge 0$$ so $$x^2+y^2+z^2\ge\frac{(ax+by+cz)^2}{a^2+b^2+c^2}=\frac{k^2}{a^2+b^2+c^2}$$

This is in fact an elemental proof of Cauchy–Schwarz_inequality on $\mathbb{R}^3$

Answer 2

The normal of the $ax+by+cz=k$ $=\bigtriangledown(ax+by+cz)= a\hat{i}+b\hat{j}+c\hat{k}=\vec{n_1}$

This magnitude of $\vec{n_1}$ is $\sqrt{a^2+b^2+c^2}$

similarly finding the normal for $x^2+y^2+z^2$ we get $\vec{n_2}=2x\hat{i}+2y\hat{j}+2z\hat{k}$

Magnitude of $\vec{n_2}$ is $2\sqrt{x^2+y^2+z^2}$

Applying Cauchy Schwartz inequality for $\vec{n_1}$ and $\vec{n_2}$

$\vec{n_1}\cdot\vec{n_1} \leq |\vec{n_1}||\vec{n_2}|$

The minimum is got when $\vec{n_1}$ is parallel (or overlapped) with $\vec{n_2}$ that is the above inequality becomes

$\vec{n_1}\cdot\vec{n_1} = |\vec{n_1}||\vec{n_2}|$

Solving this we get the required answer