f-x-1-1-2-2x-2x-x-4-prove-that-f-x-3-8-

Question Number 79588 by loveineq. last updated on 26/Jan/20

f (x) = \frac{1}{1 + 2 \sqrt{2 x} + 2 x} + \frac{x}{4}

prove that f (x) ⩾ \frac{3}{8} .

Commented by MJS last updated on 26/Jan/20

2 \sqrt{2 x} or 2 \sqrt{2} x ?

Commented by loveineq. last updated on 26/Jan/20

is 2 \sqrt{2 x} not 2 \sqrt{2} x

Commented by john santu last updated on 26/Jan/20

let \sqrt{2 x} = t \Rightarrow 2 x = t^{2}

f (x) = \frac{1}{1 + 2 t + t^{2}} + \frac{t^{2}}{8}

f (x) = \frac{1}{{(t + 1)}^{2}} + \frac{t^{2}}{8} = {(t + 1)}^{- 2} + \frac{t^{2}}{8}

f^{'} (x) = {- 2 {(t + 1)}^{- 3} + \frac{t}{4}} \times \frac{1}{t} = 0

\frac{- 2}{{(t + 1)}^{3}} + \frac{t}{4} = 0 \Rightarrow t {(t + 1)}^{3} = 8

t = 1 \Rightarrow x = \frac{1}{2} \Rightarrow f_{\min} =

\frac{1}{1 + 2 \sqrt{2 . \frac{1}{2}} + 2 . \frac{1}{2}} + \frac{1}{8}

= \frac{1}{1 + 2 + 1} + \frac{1}{8} = \frac{3}{8}

Answered by MJS last updated on 26/Jan/20

\frac{1}{2 x + 2 \sqrt{2} x^{\frac{1}{2}} + 1} + \frac{x}{4} < \frac{3}{8}

\Rightarrow

x^{2} + \sqrt{2} x^{\frac{3}{2}} - x - \frac{3 \sqrt{2}}{2} x^{\frac{1}{2}} + \frac{5}{4} < 0

{(\sqrt{x})}^{4} + \sqrt{2} {(\sqrt{x})}^{3} - {(\sqrt{x})}^{2} - \frac{3 \sqrt{2}}{2} \sqrt{x} + \frac{5}{4} < 0

{(\sqrt{x} - \frac{\sqrt{2}}{2})}^{2} ({(\sqrt{x})}^{2} + 2 \sqrt{2} \sqrt{x} + \frac{5}{2}) < 0

wrong \Rightarrow \frac{1}{2 x + 2 \sqrt{2} x^{\frac{1}{2}} + 1} + \frac{x}{4} ⩾ \frac{3}{8}

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