For $a,b,c,d \in \mathbb{R}$, is $\sqrt{a^2+b^2} + \sqrt{c^2+d^2} \geq \sqrt{(a+c)^2 + (b+d)^2}$? I have tried to show this using the fact that $\sqrt{x+y}\leq\sqrt{x}+\sqrt{y}\; $ for non-negative $x,y$ but haven't been able to arrive at a proof.
Simple algebra square root inequality
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linear-algebra
algebra-precalculus
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0Hm, the most immediate first step would be to square both sides. – 2017-01-26
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0Yes, that gives $\sqrt{(a^2 + b^2)(c^2 + d^2)} \geq? ac + bd$ but where to proceed from there? – 2017-01-26
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0try some examples is the next step. Many examples. – 2017-01-26
3 Answers
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Consider the vectors $x=\langle{a,b}\rangle$ and $y=\langle{c,d}\rangle$ By the triangle inequality, we have that $$\sqrt{a^2+b^2} + \sqrt{c^2+d^2}=||x||+||y|| \ge ||x+y|| =\sqrt{(a+c)^2 + (b+d)^2}$$ So we have $$\sqrt{a^2+b^2} + \sqrt{c^2+d^2} \ge \sqrt{(a+c)^2 + (b+d)^2}$$
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Squaring both sides gives $\sqrt{(a^2+b^2)(c^2+d^2)}\geq ac+bd$ you already have. Another squaring gives $a^2d^2+b^2c^2\geq 2abcd$, which has $x^2+y^2\geq 2xy$ structure, which holds.
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Consider the vectors $v_1=\langle{a,b}\rangle$ and $v_2=\langle{c,d}\rangle$. Recall that the dot product can be written as $$v_1 \cdot v_2 = |v_1||v_2|\cos\theta\leq|v_1||v_2|.$$ Then we substitute: $$v_1 \cdot v_2 = ac+bd\leq|v_1||v_2|=\sqrt{a^2+b^2}\sqrt{c^2+d^2}.$$ This proves the simplification you made, and therefore proves that $$\sqrt{a^2+b^2} + \sqrt{c^2+d^2} \geq \sqrt{(a+c)^2 + (b+d)^2}. \quad\quad\square$$