We are tasked with proving the inequality

$\log_2 a + \log_2 b > \log_2(ab)$

using the Cauchy-Schwarz inequality.

Step 1: Simplifying the Expression

First, recall the basic logarithmic identity:

$\log_b(xy) = \log_b(x) + \log_b(y)$

Applying this identity to the right-hand side:

$\log_2(ab) = \log_2 a + \log_2 b$

So, the inequality becomes:

$\log_2 a + \log_2 b > \log_2 a + \log_2 b$

This clearly doesn't hold, as it simplifies to $\log_2 a + \log_2 b > \log_2 a + \log_2 b$, which is never true.

Conclusion

The inequality is false, and no further proof with Cauchy-Schwarz or any other technique would change this result, because the two sides are equal.

---

Would you like to explore any other inequalities or discuss any further steps?

Here are 5 related questions:

1. How can the Cauchy-Schwarz inequality be used in other types of logarithmic inequalities?
2. What is the general form of the Cauchy-Schwarz inequality in vector space?
3. Can we prove a different logarithmic inequality using Cauchy-Schwarz?
4. How does the Cauchy-Schwarz inequality help in understanding norms in vector spaces?
5. What are some classic examples of inequalities proven by Cauchy-Schwarz?

Tip: Always verify that both sides of an inequality are valid and equivalent before trying to use advanced techniques like Cauchy-Schwarz!