# Show that $|z_1 + z_2|^2 < (1+C)|z_1|^2 + \left(1 + \frac{1}{C}\right) |z_2|^2$

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.

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

## 2 Answers

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

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

• I don't understand this .. Can you please explain Jan 3, 2016 at 13:42
• @Tejus: Just multiply both inequalities out and check that they're equivalent. Jan 3, 2016 at 13:45
• Got it!!! thanksss Jan 3, 2016 at 13:45

Hint: $$\lvert z_1+ z_2\rvert^2 +\lvert z_1 - z_2\rvert^2=2\lvert z_1\rvert^2 +2\lvert z_2\rvert^2.$$