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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.

April 9, 2017

## Graphing inequality (Tomato subjective 90)

© Cheenta 2017

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