Question Number 145191 by mathdanisur last updated on 03/Jul/21
$${if}\:\:{q}\geqslant\mathrm{1}\:\:{and}\:\:{x}>−\mathrm{1}\:\:{then}: \\ $$$$\left(\mathrm{1}+{x}\right)^{\boldsymbol{{q}}} \:\geqslant\:\left(\mathrm{1}+{x}\right)^{\boldsymbol{{q}}−\mathrm{1}} \:+\:{x}\:\geqslant\:\mathrm{1}+{qx} \\ $$
Answered by Olaf_Thorendsen last updated on 03/Jul/21
$$\left(\mathrm{1}+{x}\right)^{{q}} \:=\:\left(\mathrm{1}+{x}\right)^{{q}−\mathrm{1}} \left(\mathrm{1}+{x}\right) \\ $$$$=\:\left(\mathrm{1}+{x}\right)^{{q}−\mathrm{1}} +{x}\left(\mathrm{1}+{x}\right)^{{q}−\mathrm{1}} \\ $$$$\geqslant\:\left(\mathrm{1}+{x}\right)^{{q}−\mathrm{1}} +{x} \\ $$$$\geqslant\:\mathrm{1}+\left({q}−\mathrm{1}\right){x}+{x}\:=\:\mathrm{1}+{qx} \\ $$
Commented by mathdanisur last updated on 03/Jul/21
$${thank}\:{you}\:{Ser},\:{cool} \\ $$