Question Number 206355 by mr W last updated on 12/Apr/24
$${if}\:{the}\:{sum}\:{of}\:{three}\:{positive}\:{real}\: \\ $$$${numbers}\:{is}\:{equal}\:{to}\:{their}\:{product}, \\ $$$${prove}\:{that}\:{at}\:{least}\:{one}\:{of}\:{the}\: \\ $$$${numbers}\:{is}\:{larger}\:{than}\:\mathrm{1}.\mathrm{7}. \\ $$
Answered by A5T last updated on 12/Apr/24
$${a}+{b}+{c}={abc}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}\Rightarrow\left({abc}\right)\geqslant\mathrm{3}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${One}\:{must}\:{be}\:{at}\:{least}\:\sqrt[{\mathrm{3}}]{\mathrm{3}^{\frac{\mathrm{3}}{\mathrm{2}}} }=\sqrt{\mathrm{3}}\approx\mathrm{1}.\mathrm{732} \\ $$
Answered by mr W last updated on 12/Apr/24
$${s}={a}+{b}+{c}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}=\mathrm{3}\sqrt[{\mathrm{3}}]{{a}+{b}+{c}}=\mathrm{3}\sqrt[{\mathrm{3}}]{{s}} \\ $$$${s}^{\frac{\mathrm{2}}{\mathrm{3}}} \geqslant\mathrm{3}\:\Rightarrow{s}\geqslant\sqrt{\mathrm{3}^{\mathrm{3}} }=\sqrt{\mathrm{27}} \\ $$$${at}\:{least}\:{one}\:{number}\:{must}\:{be}\: \\ $$$$\geqslant\frac{{s}}{\mathrm{3}}\geqslant\sqrt{\mathrm{3}}>\mathrm{1}.\mathrm{7} \\ $$