f-x-1-1-x-1-1-a-ax-ax-8-a-gt-0-x-gt-0-prove-1-lt-f-x-lt-2-

Question Number 204640 by liuxinnan last updated on 24/Feb/24

f (x) = \frac{1}{\sqrt{1 + x}} + \frac{1}{\sqrt{1 + a}} + \sqrt{\frac{a x}{a x + 8}}

a > 0 x > 0

p r o v e 1 < f (x) < 2

Answered by lepuissantcedricjunior last updated on 26/Feb/24

x > 0 a > 0

x < x + 1; a < a + 1 => 0 < \frac{x}{x + 1} < x; 0 < \frac{a}{1 + a} < a

=> 0 < \frac{1}{\sqrt{1 + x}} < 1; 0 < \frac{1}{\sqrt{1 + a}} < 1 (1)

0 < a x < ax + 8

=> 0 < \frac{ax}{ax + 8} < 1 => 0 < \sqrt{\frac{a x}{a x + 8}} < 1 (2)

=> (1) + (2)

=> 0 + 0 + 0 < \frac{1}{\sqrt{1 + x}} + \frac{1}{\sqrt{1 + a}} + \sqrt{\frac{ax}{ax + 8}} < 1 + 1 + 1

=> 0 < \frac{1}{\sqrt{1 + x}} + \frac{1}{\sqrt{1 + a}} + \sqrt{\frac{ax}{ax + 8}} < 3

=> 0 < 1 < \frac{1}{\sqrt{1 + x}} + \frac{1}{\sqrt{1 + a}} + \sqrt{\frac{ax}{ax + 8}} < 2 < 3

1 < \frac{1}{\sqrt{1 + x}} + \frac{1}{\sqrt{1 + a}} + \sqrt{\frac{ax}{ax + 8}} < 2

=> 1 < f (x) < 2

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