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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

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

Rodrigo de Azevedo
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2 Answers2

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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 $\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c}.$

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juantheron
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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

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

Rodrigo de Azevedo
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  • +2 for elegance. Can please you explain how the closed form solution is calculated? – Alt Sep 01 '17 at 17:47
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    @Alt Using the Lagrangian $$\mathcal L (\mathrm x, \lambda) := \frac 12 | \mathrm x |_2^2 + \lambda ( \mathrm a^\top \mathrm x - 1 )$$ The least-norm solution of $\rm A x = b$ is well-known, provided that $\rm A$ has full row rank. If $\mathrm a \neq 0_n$, then $\mathrm a^\top$ has full row rank. – Rodrigo de Azevedo Sep 01 '17 at 18:01