Prove that $|a|+|b|\le \sqrt{2}|z|$

Question

I was solving maths and got struk on this question.might you help me with this one.

If z=a+ib Then, prove that

$|a|+|b|\le \sqrt{2}|z|$

I don't know how to start it.

Help me.

Answer 1

Use $$\frac{x+y}{2}\le\sqrt{\frac{x^2+y^2}{2}}$$

Answer 2

Since $|z|=\sqrt{a^2+b^2} $ so the inequality is equivalent (taking the square) $$(|a|+|b|)^2\le 2(a^2+b^2)\iff 0\le a^2+b^2-2|ab|=(|a|-|b|)^2$$ which is true.

Answer 3

By convexity, $$\sqrt 2|z|=\sqrt{2a^2+2b^2}=\sqrt{\frac{4a^2+4b^2}{2}}\ge \frac{\sqrt{4a^2}+\sqrt{4b^2}}{2}=|a|+|b|$$