@bemath@ log _(∣2x−(1/2)∣) (x+1+(1/x))≥log _(∣2x−(1/2)∣) (x^2 +1+(1/x^2 ))


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Question Number 106996 by bemath last updated on 10/Aug/20

$@ b e m a t h @$ $\log_{∣ 2 x - \frac{1}{2} ∣} (x + 1 + \frac{1}{x}) ⩾ \log_{∣ 2 x - \frac{1}{2} ∣} (x^{2} + 1 + \frac{1}{x^{2}})$
Commented by bemath last updated on 10/Aug/20

$t h a n k y o u$
	Answered by bobhans last updated on 10/Aug/20
	$∔bobhans∔ log _(∣2x−(1/2)∣) (x+1+(1/x)) ≥ log _(∣2x−(1/2)∣) (x^2 +1+(1/x^2 )) { ((x+1+(1/x)>0)),((x^2 +1+(1/x^2 )>0)),((∣2x−(1/2)∣>0)) :} ⇒ { ((x>0)),((x≠0)),((x≠(1/4))) :} ⇒((x^3 +x−x^4 −1)/((2x−(3/2))(2x+(1/2)))) ≥ 0 ...(×(−1) ⇔ (((x+1+(1/x))−(x^2 +1+(1/x^2 )))/(∣2x−(1/2)∣−1)) ≥ 0...(×x^2 ) ⇒((x^3 +x−x^4 −1)/((2x−(3/2))(2x+(1/2)))) ≥ 0 ...(×(−1) ⇒((x^4 −x^3 −x+1)/((2x−(3/2))(2x+(1/2)))) ≤ 0...(×(2x+(1/2))) ⇒(((x−1)(x^3 −1))/((2x−(3/2)))) ≤ 0 ⇒(((x−1)^2 (x^2 +x+1))/((2x−(3/2)))) ≤0 ; x^2 +x+1 >0 ∀x∈R ⇒(((x−1)^2 )/((2x−(3/2)))) ≤ 0 ⇒solution x∈(0, (1/4)) ∪ ((1/4),(3/4)) ∪ {1}$
	$∔ bobhans ∔$ $\log_{∣ 2 x - \frac{1}{2} ∣} (x + 1 + \frac{1}{x}) ⩾ \log_{∣ 2 x - \frac{1}{2} ∣} (x^{2} + 1 + \frac{1}{x^{2}})$ ${\begin{cases} x + 1 + \frac{1}{x} > 0 \\ x^{2} + 1 + \frac{1}{x^{2}} > 0 \\ ∣ 2 x - \frac{1}{2} ∣> 0 \end{cases} \Rightarrow {\begin{cases} x > 0 \\ x \neq 0 \\ x \neq \frac{1}{4} \end{cases}$ $\Rightarrow \frac{x^{3} + x - x^{4} - 1}{(2 x - \frac{3}{2}) (2 x + \frac{1}{2})} ⩾ 0 . . . (\times (- 1)$ $\Leftrightarrow \frac{(x + 1 + \frac{1}{x}) - (x^{2} + 1 + \frac{1}{x^{2}})}{∣ 2 x - \frac{1}{2} ∣ - 1} ⩾ 0 . . . (\times x^{2})$ $\Rightarrow \frac{x^{3} + x - x^{4} - 1}{(2 x - \frac{3}{2}) (2 x + \frac{1}{2})} ⩾ 0 . . . (\times (- 1)$ $\Rightarrow \frac{x^{4} - x^{3} - x + 1}{(2 x - \frac{3}{2}) (2 x + \frac{1}{2})} ⩽ 0 . . . (\times (2 x + \frac{1}{2}))$ $\Rightarrow \frac{(x - 1) (x^{3} - 1)}{(2 x - \frac{3}{2})} ⩽ 0$ $\Rightarrow \frac{{(x - 1)}^{2} (x^{2} + x + 1)}{(2 x - \frac{3}{2})} ⩽ 0; x^{2} + x + 1 > 0 \forall x \in R$ $\Rightarrow \frac{{(x - 1)}^{2}}{(2 x - \frac{3}{2})} ⩽ 0 \Rightarrow solution$ $x \in (0, \frac{1}{4}) \cup (\frac{1}{4}, \frac{3}{4}) \cup {1}$
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