f-x-4-4x-2-3-4x-2-4x-5-prove-that-f-x-2-

Question Number 79644 by loveineq. last updated on 26/Jan/20

f (x) = \frac{4 (4 x^{2} + 3)}{4 x^{2} + 4 x + 5}

prove that f (x) ⩾ 2 .

Commented by john santu last updated on 27/Jan/20

let f (x) < 2

\frac{{16 x}^{2} + 12}{{4 x}^{2} + 4 x + 5} < 2 \Rightarrow {4 x}^{2} + 4 x + 5 > 0 \forall x \in R

{16 x}^{2} + 12 < {8 x}^{2} + 8 x + 10

{8 x}^{2} - 8 x + 2 < 0 \Rightarrow Δ = 64 - 4.8.2 = 0

x^{2} - x + \frac{1}{4} = {(x - \frac{1}{2})}^{2} ⩾ 0

contradiction with the initial

statement .

so f (x) ⩾ 2 .

Answered by mr W last updated on 27/Jan/20

f (x) = \frac{4 (4 x^{2} + 3)}{4 x^{2} + 4 x + 5}

= \frac{4 (4 x^{2} + 4 x + 5 - 4 x - 2)}{4 x^{2} + 4 x + 5}

= 4 (1 - \frac{x + \frac{1}{2}}{x^{2} + x + \frac{5}{4}})

= 4 (1 - \frac{x + \frac{1}{2}}{{(x + \frac{1}{2})}^{2} + 1})

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

= 4 (1 + \frac{1}{t + \frac{1}{t}}) ⩾ 4 (1 - \frac{1}{2}) = 2

w i t h t = - x - \frac{1}{2}

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