# Proving $|w\overline{z}+\overline{w}z|\leq 2|wz|$

I want to prove $|w\overline{z}+\overline{w}z|\leq 2|wz|$.

My attempt:

$$\begin{array}{c c}|w\overline{z}+\overline{w}z| & =|(c+id)(a-ib)+(c-id)(a+ib)| \\ & =|2(ac+bd)| \\ & =\sqrt{4(ac+bd)^2} \\ & =2\sqrt{(ac+bd)^{2}}\end{array}$$

Now

$$\begin{array}{c c} 2|wz| & =2|(a+bi)(c+id)| \\ & =|2(ac-bd)+2i(ad+bc)| \\ & =2\sqrt{(ac-bd)^{2}+(ad+bc)^{2}}, \end{array}$$

but $(ac+bd)^{2}>0$ y $(ac-bd)^{2}\geq 0$ so the second term is larger.

Is my procedure right?

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Looks right to me. (Except we'd have to conclude "equal or larger" instead of just larger. Which is what you're aiming for.) – anon May 17 '12 at 0:53
Your argument looks fine. But how do you prove that $(ac+bd)^2 \leq (ac-bd)^2 + (ad+bc)^2$? (Open the brackets and simplify to get a square $\geq 0$.) An easier way is to use triangle inequality and make use of the fact that $\lvert ab \rvert = \lvert a \rvert \lvert b \rvert$ and $\lvert \bar{a} \rvert = \lvert a \rvert$ in that order. – user17762 May 17 '12 at 1:09
One question,2acbd larger or equal to $(ad)^{2}+(bc)^2$ – Daniela del Carmen May 17 '12 at 1:16
@DanieladelCarmen Note that $(ad-bc)^2 \geq 0$ and hence $(ad)^2 + (bc)^2 - 2adbc \geq 0$ and hence $(ad)^2 + (bc)^2 \geq 2acbd$. – user17762 May 17 '12 at 1:48
thanks very much – Daniela del Carmen May 17 '12 at 1:53

Your answer appears to be right, but you may enjoy a different way to go about it is by using the identity $z\overline{z}=|z|^2$. To me, this method gives a bit more of a geometric idea of what is going on. With this identity in mind: $$|w\overline{z}+\overline{w}z|^2=(w\overline{z}+\overline{w}z)\overline{(w\overline{z}+\overline{w}z)}=(w\overline{z}+\overline{w}z)^2$$
Now expand this to $$(w\overline{z})^2+2(w\overline{w}z\overline{z})+(\overline{w}z)^2$$
The middle term is $2|wz|^2$. The other two terms are conjugates, so the sum is real-valued and there may be some cancellation going on in the imaginary terms when we sum them. That is, $$(w\overline{z})^2+(\overline{w}z)^2\le2|wz|^2$$
Hence, we have $$|w\overline{z}+\overline{w}z|^2\le 4|wz|^2$$
$$|w\overline{z}+\overline{w}z|\le 2|wz|$$