let-0-lt-a-lt-b-prove-that-ln-1-a-b-ln-1-b-a-lt-ln2-2-

Question Number 80816 by ~blr237~ last updated on 06/Feb/20

l e t 0 < a < b p r o v e t h a t

l n (1 + \frac{a}{b}) l n (1 + \frac{b}{a}) < {(l n 2)}^{2}

Commented by ~blr237~ last updated on 06/Feb/20

S i r, i l o o k l i k e a t (3 - 4) c r o s s i n g y o u u z e

A ⩽ B a n d B ⩾ C \Rightarrow A ⩽ C ? ?

i f n o t, p l e a s e e x p l a i n

Commented by mind is power last updated on 07/Feb/20

m i s t a c k s o r r y

Answered by mr W last updated on 07/Feb/20

l e t x = \frac{a}{b} \neq 1

f (x) = \ln (1 + x) \times \ln (1 + \frac{1}{x})

f^{'} (x) = \frac{1}{1 + x} \ln (1 + \frac{1}{x}) - \frac{1}{x^{2}} \times \frac{1}{1 + \frac{1}{x}} \ln (1 + x)

f^{'} (x) = \frac{1}{x (1 + x)} [x \ln (1 + \frac{1}{x}) - \ln (1 + x)] = 0

x \ln (1 + \frac{1}{x}) - \ln (1 + x) = 0

\ln {(1 + \frac{1}{x})}^{x} = \ln (1 + x)

{(1 + \frac{1}{x})}^{x} = (1 + x)

\Rightarrow x = 1

f_{m a x} = f (1) = {(\ln 2)}^{2}

\Rightarrow f (x) ⩽ {(\ln 2)}^{2}

f o r x \neq 1 :

f (x) < {(\ln 2)}^{2}

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