Minimum of $x^2+y^2+z^2$ subject to $ax+by+cz=1$

Question

If $ax+by+cz=1$, what is the minimum of $x^2+y^2+z^2$?

It is obvious that we can do Lagrangian multiplier

$$W=x^2+y^2+z^2-\lambda (ax+by+cz-1)$$

Answer 1

Using the Cauchy-Schwartz Inequality:

$(x^2+y^2+z^2)\cdot (a^2+b^2+c^2)\geq (ax+by+cz)^2$ , Now Given $(ax+by+cz) = 1$

So $\displaystyle (x^2+y^2+z^2)\geq \frac{1}{(a^2+b^2+c^2)}$ and equality hold when $\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c}.$

Answer 2

We would like to find the point on the hyperplane $\{ \mathrm x \in \mathbb R^n \mid \mathrm a^\top \mathrm x = 1 \}$, where $\mathrm a \neq 0_n$, that is closest to the origin. Thus, we have a least-norm problem

$$\begin{array}{ll} \text{minimize} & \| \mathrm x \|_2^2\\ \text{subject to} & \mathrm a^\top \mathrm x = 1\end{array}$$

whose solution is

$$\mathrm x_{\text{LN}} := \mathrm a \left( \mathrm a^\top \mathrm a \right)^{-1}$$

whose squared $2$-norm is

$$\| x_{\text{LN}} \|_2^2 = \left( \mathrm a^\top \mathrm a \right)^{-1} \mathrm a^\top \mathrm a \left( \mathrm a^\top \mathrm a \right)^{-1} = \left( \mathrm a^\top \mathrm a \right)^{-1} = \color{blue}{\frac{1}{\| \mathrm a \|_2^2}}$$