TITLE: Prove a norm inequality by Lagrange multipliers
QUESTION [5 upvotes]: I would like to prove that $\left \| x \right \|_1 \le \sqrt{n} \left \| x \right \|_2$ for all $x \in \mathbb{R}^n$ using Lagrangian multipliers.
Thank you all for your help!

REPLY [6 votes]: Hint: Minimize $(\left \| x \right \|_2)^2$ under the constraint $\left \| x \right \|_1=C$ where $C$ is a constant.
Note that the optimal $x$ may either be at a point of non-differentiability of $\|x\|_1$ or where the contour of $(\|x\|_2)^2$ is tangent to $\|x\|_1=C$. The local optimum in the latter case can be found using Lagrange multipliers. In the former case show that if some $x_i=0$ then $(\|x\|_2)^2$ is not minimal (by deleting the zero components of $x$ and thereby reducing to the Lagrange Multiplier case).