For any two non-zero complex numbers ` z_(1) , z_(2)` the inequality
`(| z_(1)| + | z_(2)|)|(z_(1))/(|z_(1)|)+ (z_(2))/(|z_(2)|)| le 2 (| z_(2) | + | z_(2) | )`

Answer

To prove the inequality

\[ (|z_1| + |z_2|) \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq 2 (|z_1| + |z_2|) \]

for any two non-zero complex numbers \( z_1 \) and \( z_2 \), we can follow these steps:

### Step 1: Rewrite the Left-Hand Side

Start with the left-hand side of the inequality:

\[ LHS = (|z_1| + |z_2|) \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \]

### Step 2: Use the Triangle Inequality

By the triangle inequality, we know that:

\[ \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq \left| \frac{z_1}{|z_1|} \right| + \left| \frac{z_2}{|z_2|} \right| \]

Since \( \left| \frac{z_1}{|z_1|} \right| = 1 \) and \( \left| \frac{z_2}{|z_2|} \right| = 1 \), we have:

\[ \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq 1 + 1 = 2 \]

### Step 3: Substitute Back into the LHS

Substituting this result back into the left-hand side, we get:

\[ LHS \leq (|z_1| + |z_2|) \cdot 2 \]

### Step 4: Simplify the Expression

This simplifies to:

\[ LHS \leq 2 (|z_1| + |z_2|) \]

### Step 5: Conclusion

Thus, we have shown that:

\[ (|z_1| + |z_2|) \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq 2 (|z_1| + |z_2|) \]

This completes the proof of the inequality.

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