Proving an identity relating to the complex modulus: $z\bar{a}+\bar{z}a \leq 2|a||z|$

First, notice that one has: $$z\overline{a}+\overline{z}a=z\overline{a}+\overline{z\overline{a}}=2\textrm{Re}(z\overline{a}).$$ Now, recall that: $$\forall w\in\mathbb{C},\textrm{Re}(w)\leqslant |w|.$$ Whence the result.

First you have to convince yourself that the lhs of your inequality is a real. Or else, the exercise has no sense.

It is the case because $\overline{z\overline a+\overline za} = \overline za + z\overline a$, so your number is real.

Now all you need is the triangle inequality : $$z\overline a+\overline za \le \left|z\overline a+\overline za\right| \le \left|z\overline a\right| + \left|\overline za\right| = \left|\overline z\right|\left|a\right|+\left|z\right|\left|\overline a\right| = 2\left|z\right|\left|a\right|$$

Hints:

By a formula for the complex norm $$ 0\leq|z+a|^2=(z+a)\overline{(z+a)}=(z+a)(\overline{z}+\overline{a})=|z|^2+z\overline{a}+\overline{z}a+|a|^2 $$

By the triangle inequality: $$ |z+a|^2\leq(|z|+|a|)^2=|z|^2+2|z||a|+|a|^2 $$