Question Number 159494 by HongKing last updated on 17/Nov/21
Answered by qaz last updated on 18/Nov/21
$$\left(\mathrm{1}+\mathrm{a}\right)\left(\mathrm{1}+\frac{\mathrm{b}}{\mathrm{a}}\right)\left(\mathrm{1}+\frac{\mathrm{c}}{\mathrm{b}}\right)\left(\mathrm{1}+\frac{\mathrm{81}}{\mathrm{c}}\right) \\ $$$$\geqslant\left(\mathrm{1}+\sqrt{\mathrm{a}}\centerdot\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\frac{\mathrm{c}}{\mathrm{b}}}\centerdot\sqrt{\frac{\mathrm{81}}{\mathrm{c}}}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{1}+\sqrt{\mathrm{b}}\right)^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\frac{\mathrm{81}}{\mathrm{b}}}\right)^{\mathrm{2}} \\ $$$$\geqslant\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{b}}\centerdot\sqrt[{\mathrm{4}}]{\frac{\mathrm{81}}{\mathrm{b}}}\right)^{\mathrm{4}} \\ $$$$=\left(\mathrm{1}+\mathrm{3}\right)^{\mathrm{4}} \\ $$$$=\mathrm{256} \\ $$
Commented by HongKing last updated on 18/Nov/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{cool} \\ $$