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I know this inequality is true, but I don't know how to prove it.

$$\sum_{i=1}^n\sqrt{A_i^2+B_i^2} \geq \sqrt{\left(\sum_{i=1}^nA_i\right)^2+\left(\sum_{i=1}^nB_i\right)^2} $$

Any simple equation where N is 2 or 3 could work for me too. Thank you!

original problem image

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closed as off-topic by Martin R, Lee David Chung Lin, Gibbs, Aweygan, GNUSupporter 8964民主女神 地下教會 Feb 9 at 22:16

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  • $\begingroup$ If you click "This is an inequality that I need to prove" the formula will show... $\endgroup$ – Ujae Kang Feb 9 at 1:52
  • $\begingroup$ What steps have you tried? $\endgroup$ – Alex Feb 9 at 1:54
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    $\begingroup$ Trying squaring both sides. $\endgroup$ – lightxbulb Feb 9 at 2:19
  • $\begingroup$ Voila the "Minkowski inequality" , isn't it ? $\endgroup$ – Rohan Shinde Feb 9 at 5:36
  • $\begingroup$ @Digamma Yes, of course! Do you know to prove it? $\endgroup$ – Michael Rozenberg Feb 9 at 8:42
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||$\sum_{i=1}^nc_i$|| <= $\sum_{i=1}^n||c_i||$ where $c_i$= $(A_i,B_i)$ and || || is the Euclidean norm ; This is by induction on n from the triangle inequality for the Euclidean norm .

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For $n=1$ it's trivial.

For a vector $v=(A,B)\in \Bbb R^2$ let $\|v\|=\sqrt {A^2+B^2}.$

Let $S(n)$ be the statement $\sum_{j=1}^n\|v_j\|\ge \|\sum_{j=1}^nv_j\|$.

What you want to prove is that $S(n)$ is true for all $ n.$

Assume that you have proven $S(2).$

Now suppose $n\ge 3$ and that $S(n-1)$ is true. Let $v'_1=\sum_{j=1}^{n-1}v_j$ and let $v'_2=v_n.$ Then
$$\sum_{j=1}^n\|v_j\|=(\,\sum_{j=1}^{n-1}\|v_j\|\,)+\|v_n\|\ge$$ $$\ge \|\sum_{j=1}^{n-1}v_j\|+\|v_n\|=$$ $$=\|v'_1\|+\|v'_2\|\ge$$ $$\ge \|v'_1+v'_2\|=$$ $$=\|\sum_{j=1}^nv_j\|.$$

So if you can prove $S(2)$ then $S(n$) holds for all $n\ge 3$ by induction.

So all you have to do now is prove $S(2).$

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By C-S $$\sum_{i=1}^n\sqrt{A_i^2+B_i^2}=\sqrt{\sum_{i=1}^n(A_i^2+B_i^2)+2\sum_{1\leq i<j\leq n}\sqrt{(A_i^2+B_i^2)(A_j^2+B_j^2)}}\geq$$ $$\geq\sqrt{\sum_{i=1}^n(A_i^2+B_i^2)+2\sum_{1\leq i<j\leq n}(A_iA_j+B_iB_j)}=$$ $$=\sqrt{\sum_{i=1}^na_i^2+2\sum_{1\leq i<j\leq n}A_iA_j+\sum_{i=1}^nB_i^2+2\sum_{1\leq i<j\leq n}B_iB_j}=\sqrt{\left(\sum_{i=1}^nA_i\right)^2+\left(\sum_{i=1}^nB_i\right)^2}.$$ For $n=3$ it seems so: $$\sqrt{A_1^2+B_1^2}+\sqrt{A_2^2+B_2^2}+\sqrt{A_3^2+B_3^2}=$$ $$=\sqrt{\sum_{i=1}^3(A_i^2+B_i^2)+2\left(\sqrt{(A_1^2+B_1^2)(A_2^2+B_2^2)}+\sqrt{(A_1^2+B_1^2)(A_3^2+B_3^2)}+\sqrt{(A_2^2+B_2^2)(A_3^2+B_3^2)}\right)}\geq$$ $$\geq\sqrt{\sum_{i=1}^3(A_i^2+B_i^2)+2\left(A_1A_2+B_1B_2+A_1A_3+B_1B_3+A_2A_3+B_2B_3\right)}=$$ $$=\sqrt{A_1^2+A_2^2+A_3^2+2(A_1A_2+A_1A_3+A_2A_3)+B_1^2+B_2^2+B_3^2+2(B_1B_2+B_1B_3+B_2B_3)}=$$ $$=\sqrt{(A_1+A_2+A_3)^2+(B_1+B_2+B_3)^2}.$$

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