91Ó°ÊÓ

According to the result of Exercise 20 in Section 9.1, we have \(y^z > z\), where \(y = 1 - \frac{1}{z+1}\) and \(z = n^2\) for all integers \(n\geq 1\). Now, we replace the values of \(y\) and \(z\) in the inequality.

We have: \((1 - \frac{1}{n^2 + 1})^{n^2} > n^2\) We notice that the term inside the parenthesis equals \(\frac{n^2}{n^2+1}\), so we can simplify the inequality: \(\left(\frac{n^2}{n^2 + 1}\right)^{n^2} > n^2\)

Since \(\log_2 u < u\) for all positive real numbers, we can apply this fact to our inequality. Applying the given fact to the left side of our inequality: \(\log_2 \left(\left(\frac{n^2}{n^2 + 1}\right)^{n^2}\right) < \left(\frac{n^2}{n^2 + 1}\right)^{n^2}\) And applying the given fact to the right side as well: \(n^2 > \log_2 {(n^2)}\) Now, since the left side inequality is smaller than the left side of the right inequality, we have: \(\log_2 \left(\left(\frac{n^2}{n^2 + 1}\right)^{n^2}\right) < n^2 > \log_2 {(n^2)}\)

From Step 3, we derived two inequalities: \(\log_2 \left(\left(\frac{n^2}{n^2 + 1}\right)^{n^2}\right) < n^2\) and \(n^2 > \log_2 {(n^2)}\) To prove that \(\log_2 x < x^{\frac{1}{n}}\), we first need to connect these inequalities. Considering the left side inequality, we raise both sides to the power of \(\frac{1}{n}\): \(\log_2\left(\left(\frac{n^2}{n^2 + 1}\right)^{\frac{n^2}{n}}\right) < n\) Simplify the left side: \(\log_2\left(\left(\frac{n^2}{n^2 + 1}\right)^{n}\right) < n\) Now, we can equate \(x = \left(\frac{n^2}{n^2 + 1}\right)^{n}\) to find the solution: \(\log_2 x < n\) Since \(n > \log_2{(n^2)}\), we have: \(\log_2 x < n > \log_2 {(n^2)}\) Finally, we can simplify the inequality chain: \(\log_2 x < n > \log_2 {(2n)^{2n}}\) Which is equivalent to: \(\log_2 x < x^{\frac{1}{n}}\) for all real numbers \(x > (2n)^{2n}\) for all integers \(n \geq 1\).