I am having some trouble with one inequality used in the proof of the triangle inequality in $\mathbb{C}$. The main issue is realizing that for $z,w \in \mathbb{C}$, we have that $2 \cdot Re(z\overline{w}) \leq 2 \cdot |z \overline{w} |$. This fact has been exploited in the proof I am currently reading as well as here and also here. However if we take $z = \frac{1}{2} + \frac{1}{3}i$ and $w = \frac{1}{4} + \frac{1}{5}i$, then it follows that $z \overline{w} = (\frac{1}{2} + \frac{1}{3}i)(\frac{1}{4} - \frac{1}{5}i) = \frac{23}{120} - \frac{1}{60}i$ and so $2 \cdot Re(z \overline{w}) = \frac{23}{60}$ and $2 \cdot |z \overline{w}| = \frac{533}{7200}$ and so $2 \cdot Re(z \overline{w}) \geq 2 \cdot |z \overline{w}|$... So my question is am I merely making a mistake somewhere or are there some restrictions on which complex numbers this inequality applies to.
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You seem to have made an error computing $|\overline{z}w|$: $\frac{533}{7200}$ is $2|\overline{z}w|^2$, not $2|\overline{z}w|$. When you remember to take the square root, you get $2|\overline{z}w|=\frac{\sqrt{533}}{60}$, which is indeed slightly larger than $\frac{23}{60}$.
Eric Wofsey
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Ahh yes, my mistake. Is there an easy proof of this fact then? – Oiler Jul 12 '16 at 03:08
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It's pretty trivial by just plugging in the definitions: for any $z=x+iy$, $|z|=\sqrt{x^2+y^2}\geq\sqrt{x^2}=|x|=|\operatorname{Re} z|$. – Eric Wofsey Jul 12 '16 at 03:11
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Duh. Not enough sleep. Thanks for the help. – Oiler Jul 12 '16 at 03:19