Show that a 1 a 1 - a 2 a 2 - a 3 a 3 - - a n a n - 1 1 - 1 a 1 - a 1 a 2 - a 2 a 3 - - a n-2 a n-1-

⇒ a _ 1 a _ 1 + a _ 2 a _ 2 + a _ 3 a _ 3 + ..... a _ n a _ n = 1 1+ a _ 2 a _ 1 a _ 2 + a _ 1 a _ 3 a _ 3 + a _ 4 a _ 4 + ..... a _ n ...

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Multiplication of Monomials

Show that a ...

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Show that
$\frac{a_{1}}{a_{1}} + \frac{a_{2}}{a_{2}} + \frac{a_{3}}{a_{3}} + \dots \frac{a_{n}}{a_{n}} = \frac{1}{1} + \frac{1}{a_{1}} + \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \dots \frac{a_{n - 2}}{a_{n - 1}}$ .

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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_{n}}{a_{1} + a_{2} + . . . + a_{n - 1}}

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\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

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a_{i} > 0

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

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