Prove or disprove that $|a_1|+|a_2|+\ldots+|a_n|\leq n\sqrt{a_1^2+\ldots+a_n^2}$, by showing that $RHS-LHS\geq 0$ if possible.

Question

Prove or disprove that $$\left|a_1\right|+\left|a_2\right|+\ldots+\left|a_n\right|\leq n\sqrt{a_1^2+\ldots+a_n^2}$$

Where $a_1,\ldots,a_n\in\mathbb{R}$ and $n\in\mathbb{N}$.

EDIT: I was hoping there is a way without using a known inequality, ie to prove that $RHS-LHS\geq 0$

Answer 1

Notice that $|a_i|^2=a_i^2$. Using the Generalized Mean Inequality, we see $$\frac{|a_1|+\cdots+|a_n|}{n}\leq \sqrt{\frac{|a_1|^2+\cdots+|a_n|^2}{n}}$$ which we can rewrite to $$|a_1|+\cdots+|a_n|\leq \sqrt{n}\sqrt{a_1^2+\cdots+a_n^2}$$ and since $\sqrt{n}\leq n$ we get $$|a_1|+\cdots+|a_n|\leq n\sqrt{a_1^2+\cdots+a_n^2}$$

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Alternatively, we can use Cauchy-Schwarz: $$(|a_1|\cdot |1|+\cdots |a_n|\cdot |1|)^2\leq (|a_1|^2+\cdots |a_n|^2)(|1|^2+\cdots+|1|^2)$$ which we can also rewrite to $$|a_1|+\cdots+|a_n|\leq \sqrt{n}\sqrt{a_1^2+\cdots+a_n^2}$$ and finishing the proof the same way.

Answer 2

I think you mean $RHS=\sqrt{n(a_1^2+a_2^2+...+a_n^2)}$.

If so we have:

$RHS-LHS=\frac{\sum\limits_{1\leq i<j\leq n}\left(|a_i|-|a_j|\right)^2}{RHS+LHS}\geq0$, which you wished.