Is right the inequality: $|a|+|b| \leq 2|a+ib|$

Question

I want know if is right this inequality: $\forall a,b \in \mathbb{R}$, $$|a|+|b| \leq 2|a+ib|$$

Answer 1

If you square both sides (which you can, since both are nonegative) you get: $$a^2+2|ab|+b^2\leq 4(a^2+b^2)$$

so $$0\leq 3x^2-2xy+3y^2= (x-y)^2+2x^2+2y^2$$

which is true. ($x=|a|$ and $y=|b|$).

Answer 2

Hint: Use that $$|a+ib|=\sqrt{a^2+b^2}$$ and then square it.

Answer 3

For real numbers $a$ and $b$ we have \begin{align*} &&a^2&\le a^2+b^2\\ &\implies&\sqrt{a^2}&\le\sqrt{a^2+b^2}\\ &&|a|&\le\sqrt{a^2+b^2} \end{align*} In the same way we can prove $|b|\le\sqrt{a^2+b^2}$. Since $\sqrt{a^2+b^2}=|a+ib|$, the asked inequality follows.

Answer 4

The numbers of both sides are non-negatives. Since $(|a|+|b|)^2 \leq (2|a+ib|)^2$ (by a straightforward computation), we conclude that $$|a|+|b| \leq 2|a+ib|.$$