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Rational Function Inequality (Tomato subjective 77)

problem: For \({x > 0}\), show that \({\displaystyle{\frac{x^n – 1}{x – 1}}{\ge}{n{x^{\frac{n – 1}{2}}}}}\), where \({n}\) is a positive integer.

solution: \({\displaystyle{\frac{x^n – 1}{x – 1}}{\ge}{n{x^{\frac{n – 1}{2}}}}}\)

\({\Leftrightarrow}\) \({\displaystyle{\frac{(x – 1)(x^{n – 1} + x^{n – 2} + ……… + x + 1)}{x – 1}}}\) \({> n x^{\frac{n – 1}{2}}}\)

\({\Leftrightarrow}\) \({\displaystyle{\frac{x^{n – 1} + x^{n – 2} + ……… + x^1 + x^0}{n}}}\) \({> x^{\frac{n – 1}{2}}} (\dagger) \)

Now to prove \( ( \dagger) \) we observe:

But  \(\displaystyle{\frac{x^{n – 1} + x^{n – 2} + ……… + x^1 + x^0}{n} \\ > \{x^{n-1}\cdot x^{n-2} \cdots x^0 \}^{\frac{1}{n}} \\ = \{x^{(n-1) + \cdots 0} \}^{\frac{1}{n}} \\ =\{ x^{\frac{n(n-1)}{2}}\}^\frac{1}{n}} \\ = x^{\frac{(n-1)}{2}}\)

Now this follows directly from AM-GM inequality.

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