Question Number 151052 by mathdanisur last updated on 17/Aug/21
$$\mathrm{if}\:\:\:\mathrm{a};\mathrm{b};\mathrm{c}\:\:\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:\:\mathrm{and} \\ $$$$\frac{\mathrm{a}}{\mathrm{1}+\mathrm{a}}\:+\:\frac{\mathrm{b}}{\mathrm{1}+\mathrm{b}}\:+\:\frac{\mathrm{c}}{\mathrm{1}+\mathrm{c}}\:=\:\mathrm{1}\:\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{abc}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Answered by dumitrel last updated on 18/Aug/21
$$\mathrm{3}{r}+\mathrm{2}{q}+{p}=\mathrm{1}+{p}+{q}+{r}\Rightarrow\mathrm{2}{r}+{q}=\mathrm{1} \\ $$$${q}^{\mathrm{3}} \geqslant\mathrm{27}{r}^{\mathrm{2}} \Rightarrow\left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{3}} \geqslant\mathrm{27}{r}^{\mathrm{2}} \Rightarrow \\ $$$$\mathrm{8}{r}^{\mathrm{3}} +\mathrm{15}{r}^{\mathrm{2}} +\mathrm{6}{r}−\mathrm{1}\leqslant\mathrm{0}\Rightarrow\left(\mathrm{8}{r}−\mathrm{1}\right)\left({r}^{\mathrm{2}} +\mathrm{2}{r}+\mathrm{1}\right)\leqslant\mathrm{0}\Rightarrow{r}\leqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Commented by mathdanisur last updated on 18/Aug/21
$$\mathrm{nice}\:\mathrm{Ser}\:\mathrm{thank}\:\mathrm{you} \\ $$