Question Number 25170 by NECx last updated on 05/Dec/17
$${prove}\:{that}\:{n}^{\mathrm{2}} >{n}−\mathrm{5}\:{for}\:{integral}\: \\ $$$${n}\geqslant\mathrm{3}\: \\ $$
Commented by mrW1 last updated on 06/Dec/17
$${maybe}\:{the}\:{question}\:{is} \\ $$$${n}^{\mathrm{2}} >{n}+\mathrm{5}\:{for}\:{n}\geqslant\mathrm{3} \\ $$
Commented by jota@ last updated on 11/Dec/17
$${Yes} \\ $$
Answered by jota+ last updated on 09/Dec/17
$$\mathrm{3}^{\mathrm{2}} >\mathrm{3}−\mathrm{5} \\ $$$${k}^{\mathrm{2}} >{k}−\mathrm{5}\:\:\:\:\:\:\:\:\left(\mathrm{1}\right)\:\:\:{hipotesis} \\ $$$$\mathrm{2}{k}+\mathrm{1}>\mathrm{1}\:\:\:\:\:\:\:\left(\mathrm{2}\right)\:\:\:{if}\:\:{k}\geqslant\mathrm{3} \\ $$$${then} \\ $$$$\left({k}+\mathrm{1}\right)^{\mathrm{2}} >\left({k}+\mathrm{1}\right)−\mathrm{5} \\ $$
Commented by NECx last updated on 06/Dec/17
$${thank}\:{you}\:{so}\:{much}\: \\ $$
Answered by mrW1 last updated on 05/Dec/17
$${n}^{\mathrm{2}} ={n}×{n}\geqslant\mathrm{3}{n}={n}+\mathrm{2}{n}\geqslant{n}+\mathrm{6}>{n}−\mathrm{5} \\ $$