Question

Prove that for positive sequences a_n and b_n,
\[\sqrt{{a_1}^2+{b_1}^2}+\sqrt{{a_2}^2+{b_2}^2}+\cdots+\sqrt{{a_n}^2+{b_n}^2} \geq \\ \qquad \qquad \sqrt{(a_1+a_2+\cdots+a_n)^2+(b_1+b_2+\cdots+b_n)^2}\]
My idea is to use vectors and their lengths. It's easy for n=2, but I can't generalize to an arbitrary n. Can anyone help me with this?

Answer 1

Nvm. Just realized that my induction solution went awry and after fixing this, the answer unveiled itself.