How I can prove this inequality? How can I prove it using Cauchy–Schwarz inequality?

Let $a_1,...,a_n,b_1,....,b_n$ are any real numbers.

$\sqrt {\sum_{k=1}^n (a_k+b_k)^2)} \leq \sqrt {\sum_{k=1}^n {a_k}^2} \sqrt {\sum_{k=1}^n {b_k}^2}$

  • $\begingroup$ Do you mean $\sqrt{\sum_{k=1}^n a_k+b_k}=\sqrt{\sum_{k=1}^n a_k^2}+\sqrt{\sum_{k=1}^n a_k^2}$? $\endgroup$ Mar 20, 2019 at 18:19

1 Answer 1


This is false when $a_1 = b_1= 1$ and $a_k = b_k = 0$ for $k \ne 1$.


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