nice-calculus-if-b-gt-a-gt-0-then-prove-a-b-2-gt-b-a-ln-b-ln-a-gt-ab-

Question Number 128005 by mnjuly1970 last updated on 03/Jan/21

\dots . n i c e c a l c u l u s \dots

i f b > a > 0 t h e n p r o v e ::

\frac{a + b}{2} > \frac{b - a}{l n (b) - l n (a)} > \sqrt{a b}

\dots \dots \dots \dots \dots \dots \dots \dots \dots .

Answered by mindispower last updated on 06/Jan/21

\Leftrightarrow \frac{\frac{b}{a} + 1}{2} > \frac{\frac{b}{a} - 1}{l n (\frac{b}{a})} > \sqrt{\frac{b}{a}}

\frac{b}{a} = x, x > 1

\Rightarrow \frac{x + 1}{2} = \frac{x - 1}{l n (x)} > \sqrt{x}

\Leftrightarrow l n (x) > \frac{2 (x - 1)}{x + 1} . . 1

& l n (x) < \sqrt{x} - \frac{1}{\sqrt{x}}, . . 2

1 \Leftrightarrow f (x) = l n (x) - 2 + \frac{4}{x + 1} > 0

f^{'} (x) = \frac{1}{x} - \frac{4}{{(x + 1)}^{2}} = \frac{{(x - 1)}^{2}}{x {(x + 1)}^{2}} > 0

f (x) > f (1), \forall x > 1

f (1) = l n (1) - 2 + \frac{4}{2} = 0

\Rightarrow 1 . . t r u e

2 \Leftrightarrow g (x), \sqrt{x} - \frac{1}{\sqrt{x}} - l n (x) > 0

g^{'} (x) = \frac{1}{2 \sqrt{x}} + \frac{1}{2 x \sqrt{x}} - \frac{1}{x} = \frac{x + 1 - 2 \sqrt{x}}{2 x \sqrt{x}} = \frac{{(\sqrt{x} - 1)}^{2}}{2 x \sqrt{x}}

g^{'} (x) > 0, g (1) = 0 \Rightarrow g (x) > g (1) = 0, \forall x > 1

2 . . t r u e

\Rightarrow \frac{b - a}{2} > \frac{b - a}{l n (b) - l n (a)} > \sqrt{a b}, \forall a, b b > a > 0