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**earboth** 1. Re-write the given inequality:

$\displaystyle

\frac{1}{\log_b( a)}+\log_b(a)\geq 2$

2. Since a > 1 and b > 1 the logarithms are greater than zero, that means they are positive. Multiply the inequality by $\displaystyle \log_b(a)$:

$\displaystyle 1+\left(\log_b(a)\right)^2\geq 2 \cdot \log_b(a)$

3. You'll get:

$\displaystyle \left(\log_b(a)\right)^2- 2 \cdot \log_b(a) +1 \geq 0$

$\displaystyle \left(\log_b(a)-1\right)^2\geq 0$

4. A square is allways greater or equal zero.