Let $z_{1}, z_{2}$ be complex numbers with $| z_{1} | = | z_{2} | = 1$ , prove that $| z_{1} + 1 | + | z_{2} + 1 | + | z_{1} z_{2} + 1 | \geq 2$

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Solution

To prove
$\begin{matrix} | z_{1} | = | z_{2} | = 2 | z_{1} + 1 | + | z_{2} + 1 | + | z_{1} z_{2} + 1 | \geq 2 w e k n o w t h a t, | z_{1} - (- 1) | \geq | | z_{1} | - | - 1 | | S o, | z_{1} - (- 1) | + | z_{2} - (- 1) | + | z_{1} z_{2} + 1 | \geq | | z_{1} | - 1 | + | | z_{2} | - 1 | + | z_{1} z_{2} + 1 | | z_{1} + 1 | + | z_{2} + 1 | + | z_{1} z_{2} + 1 | & \geq | 1 - 1 | + | 1 - 1 | + | z_{1} z_{2} + 1 | | z_{1} + 1 | + | z_{2} + 1 | + | z_{1} z_{2} + 1 | & \geq 0 + 0 + | z_{1} z_{2} | + 1 | z_{1} + 1 | + | z_{2} + 1 | + | z_{1} z_{2} + 1 | & \geq 0 + 0 + | z_{1} | . | z_{2} | + 1 | z_{1} + 1 | + | z_{2} + 1 | + | z_{1} z_{2} + 1 | & \geq 0 + 0 + 1.1 + 1 | z_{1} + 1 | + | z_{2} + 1 | + | z_{1} z_{2} + 1 | & \geq 2 H e n c e p r o v e d [u sin g | a + b | = | a | + | b | i f a . b \geq 0] \end{matrix}$

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