Show that $\frac{1}{2}(|z|+|w|)|\frac{z}{|z|} + \frac{w}{|w|}| \le |z+w|$ with $z \neq 0$ and $w\neq0$

Question

Let $z,w \in \mathbb{C}$ Show that $\frac{1}{2}(|z|+|w|)|\frac{z}{|z|} + \frac{w}{|w|}| \le |z+w|$ with $z \neq 0$ and $w\neq0$

I know that $|z+w|\le |z| + |w|$ And that $z \cdot\bar z = |z|^2$

I tried to change values in the inequality but can’t get anything.

According to Proving result of complex numbers I think that proving $0 \le |\frac{z}{|z|} + \frac{w}{|w|}| \le 2$Would prove the inequality. Note: I couldn’t prove this either.

Could anybody give me a hint? Thank you

Answer 1

So we have to prove: $$\Big|z+w+{|w|z\over |z|}+{|z|w\over |w|}\Big|\leq 2|z+w|$$

By triangle inequality we have $$\Big|\color{red}{z+w}+\color{blue}{{|w|z\over |z|}+{|z|w\over |w|}}\Big|\leq |\color{red}{z+w}|+\Big|\color{blue}{{|w|z\over |z|}+{|z|w\over |w|}}\Big|$$

so it is enough to check if $$\Big|{|w|z\over |z|}+{|z|w\over |w|}\Big|\leq |z+w|$$ is true.

Let $u=z/w$ then we have to prove $$\Big||u|+{u\over |u|}\Big|\leq |1+u|$$ or $$||u|^2+u|\leq |u||1+u|$$

but this is actually even equalty so the claim is proven.

Answer 2

Let $z=re^{i\theta},w=se^{i\psi},\;r,s\in\mathbb R^+,\theta,\psi\in\mathbb R$. The inequality can be rewritten as $$\frac{r+s}2\cdot|e^{i\theta}+e^{i\psi}|<|re^{i\theta}+se^{i\psi}|$$

We first work on the right side of the inequality. We note that $$|re^{i\theta}+se^{i\psi}|=\sqrt{(r\cos\theta+s\cos\psi)^2+(r\sin\theta+s\sin\psi)^2}$$$$=\sqrt{r^2+s^2+2rs\cos(\theta+\psi)}$$Then, note that $$|e^{i\theta}+e^{i\psi}|=\sqrt{2+2\cos(\theta+\psi)}$$$$\frac{r+s}2\cdot|e^{i\theta}+e^{i\psi}|=\sqrt{\frac{(r+s)^2}2\cdot(1+\cos(\theta+\psi))}$$The rest is just simple algebra. Can you take it from here?