An inequality concerning triangle inequality.

Question

$a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3 \ge 0$, and given that $a_i+b_i \ge c_i$ for $i = 1,2,3$. I'd like the following inequality to hold, but can't find a proof, so I'd appreciate some help. $$\sqrt{a_1^2 + a_2^2 + a_3^2} + \sqrt{b_1^2 + b_2^2 + b_3^2} \ge \sqrt{c_1^2 + c_2^2 + c_3^2} $$

Answer 1

Hint: Think of vectors $A(a_1, a_2, a_3)$ and $B(b_1, b_2, b_3)$. Now $\lvert A \rvert + \lvert B \rvert \ge \lvert A+B \rvert $.

Answer 2

Good news. It does hold.

Note that applying Minkowski's inequality to the LHS, we have:

$$\sqrt{a_1^2 + a_2^2 + a_3^2} +\sqrt{b_1^2 + b_2^2 + b_3^2} \ge \sqrt{(a_1 +b_1)^2+(a_2 +b_2)^2 +(a_3 +b_3)^2}$$ Now all we need to do is show that $$\sqrt{(a_1 +b_1)^2+(a_2 +b_2)^2 +(a_3 +b_3)^2} \ge \sqrt{c_1^2+c_2^2+c_3^2}$$ Which is trivial since: $$a_i+b_i \ge c_i$$