# Proving a Cauchy Schwarz Inequality [closed]

Let $a_1,a_2,\cdots,a_n ;b_1,b_2,\cdots,b_n \in \mathbb{R}^{+}$. Prove that:

$$\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2}+\cdots +\sqrt{a_n^2+b_n^2}\geq \sqrt{(a_1+a_2+\cdots+a_n)^2+(b_1+b_2+\cdots+b_n)^2 }$$

## closed as off-topic by Davide Giraudo, Leucippus, pjs36, Shailesh, Claude LeiboviciApr 18 '17 at 6:39

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• Hint: prove for $n=2$ and then use induction. – ntt Apr 17 '17 at 15:36
• It's just Minkowski – Michael Rozenberg Apr 17 '17 at 17:49

By C-S $$\sum_{i=1}^n\sqrt{a_i^2+b_i^2}=\sqrt{\sum_{i=1}^n(a_i^2+b_i^2)+2\sum_{1\leq i<j\leq n}\sqrt{(a_i^2+b_i^2)(a_j^2+b_j^2)}}\geq$$ $$\geq\sqrt{\sum_{i=1}^n(a_i^2+b_i^2)+2\sum_{1\leq i<j\leq n}(a_ia_j+b_ib_j)}=$$
$$=\sqrt{\sum_{i=1}^n a_i^2+2\sum_{1\leq i<j\leq n}a_ia_j+\sum_{i=1}^nb_i^2+2\sum_{1\leq i<j\leq n}b_ib_j}=$$ $$=\sqrt{(a_1+a_2+...+a_n)^2+(b_1+b_2+...+b_n)^2}$$ and we are done!