Question Number 156642 by MathSh last updated on 13/Oct/21
![For x∈(0;∞) - Z prove that: (({x}^3 )/([x])) + (([x]^3 )/({x})) ≥ (1/8) (x^2 + [x]^2 + {x}^2 ) [∗]-GIF and {x}=x-[x]](https://www.tinkutara.com/question/Q156642.png)
$$\mathrm{For}\:\:\mathrm{x}\in\left(\mathrm{0};\infty\right)\:-\:\mathbb{Z}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\left\{\mathrm{x}\right\}^{\mathrm{3}} }{\left[\mathrm{x}\right]}\:+\:\frac{\left[\mathrm{x}\right]^{\mathrm{3}} }{\left\{\mathrm{x}\right\}}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{8}}\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\left[\mathrm{x}\right]^{\mathrm{2}} \:+\:\left\{\mathrm{x}\right\}^{\mathrm{2}} \right) \\ $$$$\left[\ast\right]-\mathrm{GIF}\:\:\mathrm{and}\:\:\left\{\mathrm{x}\right\}=\mathrm{x}-\left[\mathrm{x}\right] \\ $$
Commented by ghimisi last updated on 14/Oct/21
![x∈(0,1)⇒[x]=0 ?????????](https://www.tinkutara.com/question/Q156685.png)
$${x}\in\left(\mathrm{0},\mathrm{1}\right)\Rightarrow\left[{x}\right]=\mathrm{0}\:????????? \\ $$$$ \\ $$
Commented by MathSh last updated on 14/Oct/21
$$\mathrm{Dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{try}\:\mathrm{for}\:\mathrm{x}>\mathrm{1} \\ $$
Commented by ghimisi last updated on 14/Oct/21
$${ok} \\ $$
Answered by ghimisi last updated on 14/Oct/21
![x>1 [x]=a≥1 {x}=b<1⇒x=a+b ⇔4(a^4 +b^4 )≥ab(a^2 +ab+b^2 )⇔ 4(1+((b/a))^4 )≥((b/a))^3 +((b/a))^2 +((b/a)) (b/a)=t<1 ⇔4t^4 +4≥t^3 +t^2 +t t<1⇒t^3 +t^2 +t<3<4<4+4t^4](https://www.tinkutara.com/question/Q156702.png)
$${x}>\mathrm{1} \\ $$$$\left[{x}\right]={a}\geqslant\mathrm{1} \\ $$$$\left\{{x}\right\}={b}<\mathrm{1}\Rightarrow{x}={a}+{b} \\ $$$$\Leftrightarrow\mathrm{4}\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} \right)\geqslant{ab}\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)\Leftrightarrow \\ $$$$\mathrm{4}\left(\mathrm{1}+\left(\frac{{b}}{{a}}\right)^{\mathrm{4}} \right)\geqslant\left(\frac{{b}}{{a}}\right)^{\mathrm{3}} +\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{b}}{{a}}\right) \\ $$$$\frac{{b}}{{a}}={t}<\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{4}{t}^{\mathrm{4}} +\mathrm{4}\geqslant{t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t} \\ $$$${t}<\mathrm{1}\Rightarrow{t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t}<\mathrm{3}<\mathrm{4}<\mathrm{4}+\mathrm{4}{t}^{\mathrm{4}} \\ $$
Commented by MathSh last updated on 14/Oct/21
$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$