Elementary inequality proof $2|tx|\leq |x^2|+|t^2|$

Question

I am trying to prove the following inequality for $x\in \mathbb{R}$ und $t\in \mathbb{C}$: $$2|tx|\leq |t^2|+|x^2|$$

I have for $t=a+bi$: $$|t^2|=|a^2-b^2+aib|=\sqrt{\left(a^2+b^2\right)^2}$$ and $$2|tx|=2\sqrt{x^2}\sqrt{a^2+b^2}$$ reaching $$2\sqrt{x^2}\sqrt{a^2+b^2}\leq\sqrt{\left(a^2+b^2\right)^2}+x^2 $$

How do I continue?

Answer 1

$$\lvert x\rvert^2+\lvert t\rvert^2-2\lvert xt\rvert=\lvert x\rvert^2+\lvert t\rvert^2-2\lvert x\rvert\lvert t\rvert=\bigl(\lvert x\rvert-\lvert t\rvert\bigr)^2\geqslant0.$$

Answer 2

Hint: Write $$0\le |x|^2+|t|^2-2|tx|$$ it is $$0\le (|x|-|t|)^2$$