Basic question on inner product and norms

Question

If $x = (f,g)$ is a vector in $\mathbb{C}^2$. such that $\|x\|^2 = |f|^2 + |g|^2 = 1$ Define the operator $T: \mathbb{C}^2 \to\mathbb{C}^2$ as $Tx = (g,0)$, then $\langle Tx, x\rangle = g\bar{f}$. Why does this $|\langle Tx,x \rangle| = |g| |f|\leq \frac{1}{2} (|f|^2 + |g|^2)$ hold?

Answer 1

$\dfrac{|f|^2 + |g|^2}{2} - |g||f| = \dfrac{\left(|f| - |g|\right)^2}{2} \geq 0$

Answer 2

For all complex numbers $a$ and $b$ we have

$$\left(|a|-|b|\right)^2\ge0\iff |a||b|\le\frac12\left(|a|^2+|b|^2\right)$$