# For $z, w \in\mathbb C$, prove that $2|zw| \le |z|^2+|w|^2$

For $z, w \in \mathbb C$, prove that $2|zw| \le |z|^2+|w|^2$, and deduce that $|z+w|^2 \le 2(|z|^2+|w|^2)$, where $|v|$ is the modulus of vector $v \in \mathbb C$.

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Eugh, again? This is like the ninth question from this guy... –  Ｊ. Ｍ. Nov 4 '10 at 10:31
And I don't think any of those questions have had the right tag either... –  Hans Lundmark Nov 4 '10 at 10:33
Well, the question is clear. But this user seems to spam a lot with his homework questions -> -1 –  Djaian Nov 4 '10 at 10:36
well i try my best to behave nicely. Maybe getting some hint for what is problem in the question would help. –  laovultai Nov 4 '10 at 11:48
first, don't say "prove !", but "How to prove ?" People don't like receiving orders, but they like to answer question in order to show how smart they are. Second, if this is homework, mention it. Third, maybe some context would be nice with such question "I am following a course on <subject> and I don't understand this" or "I am reading a book where this exercise appears", and finally explain what you already tried. –  Djaian Nov 4 '10 at 12:01
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## 1 Answer

The first inequality follows from expanding $$(|z|-|w|)^2.$$ The second inequality follows from expanding $$(z+w)(\bar{z}+\bar{w}).$$ Note that $\bar{z}w+z\bar{w}=2\Re(\bar{z}w)$ and $\Re(c)\le |c|$. Then apply the first part.

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