Question Number 155621 by mathdanisur last updated on 02/Oct/21
Answered by ghimisi last updated on 03/Oct/21
$$\frac{{a}+{b}+{c}}{\mathrm{6}}=\mathrm{1}\Rightarrow \\ $$$$\left(\frac{{a}+{d}}{{a}}\right)^{\frac{{a}}{\mathrm{6}}} \left(\frac{{b}+{e}}{{b}}\right)^{\frac{{b}}{\mathrm{6}}} \left(\frac{{c}+{f}}{{c}}\right)^{\frac{{c}}{\mathrm{6}}} \leqslant\frac{{a}}{\mathrm{6}}\centerdot\frac{{a}+{d}}{{a}}+\frac{{b}}{\mathrm{6}}\centerdot\frac{{b}+{e}}{{b}}+\frac{{c}}{\mathrm{6}}\centerdot\frac{{c}+{f}}{{c}}=\mathrm{2}\Rightarrow \\ $$$$\left(\frac{{a}+{d}}{{a}}\right)^{{a}} \left(\frac{{b}+{c}}{{b}}\right)^{{b}} \left(\frac{{c}+{f}}{{c}}\right)^{{c}} \leqslant\mathrm{64}\:\:\left(\bullet\right) \\ $$$$\frac{{d}+{e}+{f}}{\mathrm{6}}=\mathrm{1}\Rightarrow \\ $$$$\left(\frac{{a}+{d}}{{d}}\right)^{\frac{{d}}{\mathrm{6}}} \left(\frac{{b}+{e}}{{e}}\right)^{\frac{{e}}{\mathrm{6}}} \left(\frac{{c}+{f}}{{f}}\right)^{\frac{{f}}{\mathrm{6}}} \leqslant\frac{{d}}{\mathrm{6}}\centerdot\frac{{a}+{d}}{{d}}+\frac{{e}}{\mathrm{6}}\centerdot\frac{{b}+{e}}{{e}}+\frac{{f}}{\mathrm{6}}\centerdot\frac{{c}+{f}}{{f}}=\mathrm{2} \\ $$$$\left(\frac{{a}+{d}}{{d}}\right)^{{d}} \left(\frac{{b}+{c}}{{e}}\right)^{{e}} \left(\frac{{c}+{f}}{{f}}\right)^{{f}} \leqslant\mathrm{64}\:\:\left(\bullet\bullet\right) \\ $$$$\left(\bullet\right)+\left(\bullet\bullet\right)\Rightarrow\frac{\left({a}+{d}\right)^{{a}+{d}} \left({b}+{e}\right)^{{b}+{e}} \left({c}+{f}\right)^{{c}+{f}} }{{a}^{{a}} {b}^{{b}} {c}^{{c}} {d}^{{d}} {e}^{{e}} {f}^{{f}} }\leqslant\mathrm{4096} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mathdanisur last updated on 03/Oct/21
$$\mathrm{Perfect}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$