We have $\left(\frac{a}{b}\right)^{1+1}<\left(\frac{a}{b}\right)^{1}$, which simplifies to $\left(\frac{a}{b}\right)^{2}<\left(\frac{a}{b}\right)$. Since $0<a<b$, $\frac{a}{b}<1$ and squaring a number less than 1 makes it smaller, so the base case holds.
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