If $z_1, z_2$ complex numbers and $u\in(0, \frac{π}{2})$ Prove that: $\frac{|z_1|^2}{\cos^2u}+\frac{|z_2|^2}{\sin^2u}\ge|z_1|^2+|z_2|^2+2Re(z_1z_2)$ If $z_1, z_2$ are complex numbers and $u\in(0, \frac{\pi}{2})$ Prove that: $$\frac{|z_1|^2}{\cos^2u}+\frac{|z_2|^2}{\sin^2u}\ge|z_1|^2+|z_2|^2+2\text{Re}(z_1z_2)$$
I was just looking at the question above and do not know where to start. Had these been real numbers we were talking about, I would have immediately thought of using Andreescu as follows:
$$\frac{|z_1|^2}{\cos^2u}+\frac{|z_2|^2}{\sin^2u}\ge \frac{|z_1+z_2|^2}{\cos^2u+\sin^2u}=|z_1+z_2|^2$$
And this is where I get stuck. I assume that this is where imaginary numbers come into play. Could you please explain to me how to finish off this question and how to solve questions with complex numbers in general (e.g. reference to some source)?
 A: $|z_1+z_2|^2=(z_1+z_2)(\overline{z_1+z_2})=(z_1+z_2)(\overline{z_1}+\overline{z_2})=|z_1|^2+|z_2|^2+2Re{(z_1\overline{z_2})}$
A: This is equivalent to proving that for all $u \in (0,\pi/2)$,
$$|z_1|^2 \tan^2 u + |z_2|^2 \cdot \frac1{\tan^2 u} \ge 2 \text{Re}{(z_1 z_2)}$$
Which is the same thing as proving that for every $x \ge 0$,
$$|z_1|^2 x^2 + |z_2|^2 \cdot \frac1{x^2} \ge 2 \text{Re}{(z_1 z_2)}$$
Which is the same thing as proving that for every $t \ge 0$,
$$|z_1|^2 t^2 - 2\text{Re}(z_1 z_2)t + |z_2|^2 \ge 0$$
Consider the quadratic on the LHS. We have  $\Delta ' = \text{Re}(z_1 z_2)^2 - |z_1|^2 |z_2|^2$.
Now $\Delta ' \le \text{Re}(z_1 z_2)^2 + \text{Im}(z_1 z_2)^2 - |z_1|^2 |z_2|^2 = |z_1 z_2|^2 - |z_1|^2 |z_2|^2 = |z_1|^2 |z_2|^2 - |z_1|^2|z_2|^2 = 0$
And since the coefficient $|z_1|^2$ of the quadratic is non-negative, this proves the desired inequality.
A: Let $z_1=x_1+y_1i$ and $z_2=x_2+y_2i,$ where $x_1$, $x_2$, $y_1$ and $y_2$ are reals.
Thus, by C-S twice we obtain:
$$\frac{|z_1|^2}{\cos^2u}+\frac{|z_2|^2}{\sin^2u}\geq\frac{(|z_1|+|z_2|)^2}{\cos^2u+\sin^2u}=|z_1|^2+|z_2|^2+2|z_1||z_2|=$$
$$=|z_1|^2+|z_2|^2+2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}\geq|z_1|^2+|z_2|^2+2\sqrt{(x_1x_2-y_1y_2)^2}=$$
$$=|z_1|^2+|z_2|^2+2|Re(z_1z_2)|\geq|z_1|^2+|z_2|^2+2Re(z_1z_2).$$
A: By C-S
$$\frac{|z_1|^2}{cos^2u}+\frac{|z_2|^2}{sin^2u}\ge \frac{{(|z_1|+|z_2)}^2}{\cos^2 u+\sin^2 u}$$
thus  we are left to prove $|z_1||z_2|\ge Re(z_1z_2)$
which is obvious!
