Problem 765

Prove that because $|1+\bar{z} w|^{2}-|z+w|^{2}$
$$\begin{array}{l} =(1+\bar{z} w)(1+z \bar{w})-(z+w)(\bar{z}+\bar{w}) \\ =1+|\bar{z}|^{2}|w|^{2}-|z|^{2}-|w|^{2} \\ =\left(1-|\bar{z}|^{2}\right)\left(1-|w|^{2}\right) . \end{array}$$

Also, because
$$|z| \leqslant 1,|w| \leqslant 1,$$

we have
$$\left(1-|z|^{2}\right)\left(1-|w|^{2}\right) \geqslant 0 \text {. }$$

Therefore,
$$|1+\bar{z} w| \geqslant|z+w| \text {. }$$