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Let $z_1$ and $z_2$ be two complex numbers. Show that there exists $C > 0$ with $$ |z_1 + z_2|^2 < (1+C)|z_1|^2 + \left(1 + \frac{1}{C}\right) |z_2|^2. $$

I tried to simplify the L.H.S and R.H.S, SNF I was finally left to compare between a real number and a complex number I really couldn't think of anything else. Please help.

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    $\begingroup$ $z_1 = z_2 = c = 1$ gives $4 < 4$, so probably the inequality should be $\leq$ instead of $<$ or $c \neq 1$. $\endgroup$ Jan 3, 2016 at 13:44

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HINT: The given inequality can be rewritten as the obvious inequality

$$\left|\sqrt{c}z_1-\frac{1}{\sqrt{c}}z_2\right|^2\ge 0$$

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    $\begingroup$ I don't understand this .. Can you please explain $\endgroup$
    – Tejus
    Jan 3, 2016 at 13:42
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    $\begingroup$ @Tejus: Just multiply both inequalities out and check that they're equivalent. $\endgroup$
    – Matt L.
    Jan 3, 2016 at 13:45
  • $\begingroup$ Got it!!! thanksss $\endgroup$
    – Tejus
    Jan 3, 2016 at 13:45
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Hint: $$\lvert z_1+ z_2\rvert^2 +\lvert z_1 - z_2\rvert^2=2\lvert z_1\rvert^2 +2\lvert z_2\rvert^2. $$

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