Test of Mathematics Solution Subjective 77

This is a Test of Mathematics Solution Subjective 77 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

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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This is a Test of Mathematics Solution Subjective 77 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

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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