# Proving $\sum\sqrt{A_i^2+B_i^2} \geq \sqrt{\left(\sum A_i\right)^2+\left(\sum B_i\right)^2}$ [closed]

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

## 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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• If you click "This is an inequality that I need to prove" the formula will show... – Ujae Kang Feb 9 at 1:52
• What steps have you tried? – Alex Feb 9 at 1:54
• Trying squaring both sides. – lightxbulb Feb 9 at 2:19
• Voila the "Minkowski inequality" , isn't it ? – Rohan Shinde Feb 9 at 5:36
• @Digamma Yes, of course! Do you know to prove it? – Michael Rozenberg Feb 9 at 8:42

||$$\sum_{i=1}^nc_i$$|| <= $$\sum_{i=1}^n||c_i||$$ 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 .

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 $$S(2).$$

By C-S $$\sum_{i=1}^n\sqrt{A_i^2+B_i^2}=\sqrt{\sum_{i=1}^n(A_i^2+B_i^2)+2\sum_{1\leq i $$\geq\sqrt{\sum_{i=1}^n(A_i^2+B_i^2)+2\sum_{1\leq i $$=\sqrt{\sum_{i=1}^na_i^2+2\sum_{1\leq i For $$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}.$$