Question Number 212741 by mathocean1 last updated on 22/Oct/24
$${a},\:{b},{c}\:\in\:\mathbb{N}^{\ast} \\ $$$$ \\ $$$${Show}\:{that}\:{ab}<{c}\:\Rightarrow\:{a}+{b}\leqslant{c} \\ $$$$ \\ $$
Answered by A5T last updated on 22/Oct/24
$${WLOG};\:{let}\:{a}\leqslant{b} \\ $$$$\Rightarrow\mathrm{2}{a}={a}+{a}\leqslant{a}+{b}\leqslant{b}+{b}=\mathrm{2}{b}\leqslant{ab}<{c}\:{when}\:{a}\geqslant\mathrm{2} \\ $$$$\Rightarrow{a}+{b}<{c} \\ $$$${when}\:{a}=\mathrm{1};\:{ab}={b}<{c}\Rightarrow{b}+\mathrm{1}\leqslant{c}\equiv{a}+{b}\leqslant{c} \\ $$