If $|\alpha|\leq 1$ and $|\beta|\leq 1$, prove that $|\alpha+\beta|\leq |1+\overline{\alpha}\beta|$ Note $\alpha$ and $\beta$ are complex numbers and $\overline{\alpha}$ is the conjugate of $\alpha$. I've tried using variations of the triangle inequality and I couldn't find anything to work.

$|\alpha + \beta|^2 = |(x_1+x_2) + (y_1+y_2)i|^2 = (x_1+x_2)^2+(y_1+y_2)^2$
$|1+\bar{\alpha}\beta|^2 = |1+(x_1-y_1i)(x_2+y_2i)|^2 = |1+x_1x_2+y_1y_2 + (x_1y_2-x_2y_1)i|^2 = (1+x_1x_2+y_1y_2)^2 + (x_1y_2-x_2y_1)^2$. Thus we prove:
$x_1^2+x_2^2+2x_1x_2 + y_1^2+y_2^2+2y_1y_2 \leq 1+x_1^2x_2^2+y_1^2y_2^2 + 2x_1x_2 + 2y_1y_2 + 2x_1y_1x_2y_2 + x_1^2y_2^2 - 2x_1y_1x_2y_2 + x_2^2y_1^2$ or:
$x_1^2 + x_2^2 + y_1^2 + y_2^2 \leq 1 + x_1^2x_2^2 + y_1^2y_2^2 + x_1^2y_2^2 + x_2^2y_1^2$
or:
$x_1^2 + x_2^2 + y_1^2 + y_2^2 \leq 1 + (x_1^2+x_2^2)(y_1^2+y_2^2)$
or:
$(x_1^2+x_2^2 - 1)(y_1^2+y_2^2 - 1) \geq 0$
which is true since $|\alpha|, |\beta| \leq 1$.
