Question Number 173303 by dragan91 last updated on 09/Jul/22
Answered by Frix last updated on 09/Jul/22
$$\mathrm{5}{a}^{\mathrm{6}} +\mathrm{15}{a}^{\mathrm{4}} +\mathrm{1}\geqslant\mathrm{16}{a}^{\mathrm{5}} +\mathrm{5}{a}^{\mathrm{2}} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{5}{a}^{\mathrm{6}} −\mathrm{16}{a}^{\mathrm{5}} +\mathrm{15}{a}^{\mathrm{4}} −\mathrm{5}{a}^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$$\left({a}−\mathrm{1}\right)^{\mathrm{4}} \left(\mathrm{5}{a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\left({a}−\mathrm{1}\right)^{\mathrm{4}} \geqslant\mathrm{0}\wedge\left(\mathrm{5}{a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{1}\right)\geqslant\frac{\mathrm{1}}{\mathrm{5}}>\mathrm{0} \\ $$$${A}\geqslant\mathrm{0}\wedge{B}>\mathrm{0}\:\Rightarrow\:{AB}\geqslant\mathrm{0} \\ $$$$\mathrm{proven}. \\ $$