Question Number 12173 by Mr Chheang Chantria last updated on 15/Apr/17
![Prove that ∀x∈[1,2] ⇒ 1−x^2 ≤ x](https://www.tinkutara.com/question/Q12173.png)
$$\boldsymbol{{Prove}}\:\boldsymbol{{that}}\:\forall\boldsymbol{{x}}\in\left[\mathrm{1},\mathrm{2}\right] \\ $$$$\Rightarrow\:\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} \:\leqslant\:\boldsymbol{{x}} \\ $$
Answered by ajfour last updated on 18/Apr/17
$${x}\geqslant\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{1}−{x}\:\leqslant\:\mathrm{0}\:\leqslant\:\:{x}^{\mathrm{2}} \\ $$$${so},\:\:\mathrm{1}−{x}^{\mathrm{2}} \:\leqslant\:{x}\:\:{for}\:{x}\in\:\left[\mathrm{1},\infty\right) \\ $$
Commented by Mr Chheang Chantria last updated on 17/Apr/17
$$\mathrm{your}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{not}\:\mathrm{enough}\: \\ $$
Answered by ajfour last updated on 17/Apr/17
![let 1−x^2 ≥ x ⇒ x^2 +x ≤ 1 ⇒ x^2 +x+(1/4) ≤ (5/4) or (x+(1/2))^2 ≤( ((√5)/2))^2 ⇒ −((√5)/2) ≤ x+(1/2) ≤ ((√5)/2) ⇒ −((((√5)+1))/2) ≤ x ≤ ((((√5)−1))/2) < 1 < 2 so if x∈ [1,2] 1−x^2 < x .](https://www.tinkutara.com/question/Q12277.png)
$${let}\:\mathrm{1}−{x}^{\mathrm{2}} \:\geqslant\:{x} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +{x}\:\leqslant\:\mathrm{1} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +{x}+\frac{\mathrm{1}}{\mathrm{4}}\:\leqslant\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${or}\:\:\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\left(\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\leqslant\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\: \\ $$$$\Rightarrow\:\:\:−\frac{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}{\mathrm{2}}\:\leqslant\:{x}\:\leqslant\:\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}}\:<\:\mathrm{1}\:<\:\mathrm{2} \\ $$$${so}\:{if}\:{x}\in\:\left[\mathrm{1},\mathrm{2}\right]\:\:\:\:\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} \:<\:\boldsymbol{{x}}\:. \\ $$