Show that a1a2 - a2a3 - a3a4 - - - an - 1an - ana1 - n- where a1- a2- - an are different positive integers-

We know that #160; AM≥ GM AM= Arithmetic mean GM= Geometric meanNow, a _ 1 a _ 2 + a _ 2 a _ 3 + a _ 3 a _ 4 +.......+ a _ n a _ 1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Geometric Mean

Show that a...

Question

Show that $\frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \frac{a_{3}}{a_{4}} + . . . . . + \frac{a_{n - 1}}{a_{n}} + \frac{a_{n}}{a_{1}} > n$ , where $a_{1}, a_{2}, . . ., a_{n}$ are different positive integers.

Open in App

Solution

Suggest Corrections

Similar questions

a_{1}, a_{2}, a_{3}, . . . a_{n}

\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . a_{n}}, \frac{a_{3}}{a_{1} + a_{2} + a_{4} + . . . + a_{n}}, \frac{a_{n}}{a_{1} + a_{2} + . . . + a_{n}}

\frac{a_{1}}{a_{1}} + \frac{a_{2}}{a_{2}} + \frac{a_{3}}{a_{3}} + \dots \frac{a_{n}}{a_{n}} = \frac{1}{1} + \frac{1}{a_{1}} + \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \dots \frac{a_{n - 2}}{a_{n - 1}}

a_{1}, a_{2}, a_{3}, . . ., a_{n}

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + . . . \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}

a_{1}, a_{2}, a_{3}, . . . . . . a_{n}

a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} . . . . + a_{n - 1} a_{n} =

a_{r} > 0; \forall r, n \in N

a_{1}, a_{2}, a_{3}, . . . . . a_{2 n}

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} =

Join BYJU'S Learning Program