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Prove that 1...

Question

Prove that
$\frac{1}{x + 1} - \frac{1}{x} \leq \frac{1}{x - 1} - \frac{1}{x - 2}$

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Solution

Solving L.H.S,

$\frac{1}{1 + x} - \frac{1}{x}$

$= \frac{x - x - 1}{x (x + 1)}$

$= \frac{- 1}{x (x + 1)}$

Solving R.H.S,

$\frac{1}{x - 1} - \frac{1}{x - 2}$

$= \frac{x - 2 - x + 1}{(x - 1) (x - 2)}$

$= \frac{- 1}{(x - 1) (x - 2)}$

It is obvious that L.H.S is $\leq$ R.H.S

That is $\frac{- 1}{x (x + 1)} \leq \frac{- 1}{(x - 1) (x - 2)}$

or
$⟹ \frac{1}{1 + x} - \frac{1}{x} \leq \frac{1}{x - 1} - \frac{1}{x - 2}$

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