$\sum \sqrt{a_n^2 + b_n^2} \geqslant \sqrt{\left(\sum a_n\right)^{2} + \left(\sum b_n\right)^{2}}$.
Edited by tramyvodoi, 26-01-2013 - 16:35.
Edited by tramyvodoi, 26-01-2013 - 16:35.
Cho $a_{1}$ $,$ $a_{2}$ $,$ $...$ $a_{n}$ $>$ $0$ $;$ $b_{1}$ $,$ $b_{2}$ $,$ $...$ $b_{n}$ $>$ $0$. Chứng minh rằng :
$\sum \sqrt{a_n^2 + b_n^2} \geqslant \sqrt{\left(\sum a_n\right)^{2} + \left(\sum b_n\right)^{2}}$.
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