Question Number 142379 by Rexzie last updated on 30/May/21
$${Show}\:{that}\:\mathrm{1}+\mathrm{3}{n}<{n}^{\mathrm{2}} \:{for}\:{every}\:{positive}\:{integer}\:{n}\geqslant\mathrm{4} \\ $$
Commented by mr W last updated on 30/May/21
$${n}\geqslant\mathrm{4} \\ $$$${n}^{\mathrm{2}} \geqslant\mathrm{4}{n}=\mathrm{3}{n}+{n}\geqslant\mathrm{3}{n}+\mathrm{4}>\mathrm{3}{n}+\mathrm{1} \\ $$