Question Number 24469 by Tinkutara last updated on 18/Nov/17
$$\mathrm{Let}\:\mathrm{2}{x}\:+\:\mathrm{3}{y}\:+\:\mathrm{4}{z}\:=\:\mathrm{9},\:{x},\:{y},\:{z}\:>\:\mathrm{0}\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{1}\:+\:{x}\right)^{\mathrm{2}} \:\left(\mathrm{2}\:+\:{y}\right)^{\mathrm{3}} \\ $$$$\left(\mathrm{4}\:+\:{z}\right)^{\mathrm{4}} \:\mathrm{is} \\ $$
Commented by sushmitak last updated on 18/Nov/17
$$\mathrm{AM}>\mathrm{GM} \\ $$$$\frac{\left(\mathrm{1}+{x}\right)×\mathrm{2}+\left(\mathrm{2}+{y}\right)×\mathrm{3}+\left(\mathrm{4}+{z}\right)×\mathrm{4}}{\mathrm{9}}\geqslant\left(\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{2}+{y}\right)^{\mathrm{3}} \left(\mathrm{4}+{z}\right)^{\mathrm{4}} \right)^{\mathrm{1}/\mathrm{9}} \\ $$$$\frac{\mathrm{24}+\mathrm{9}}{\mathrm{9}}\geqslant\left(\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{2}+{y}\right)^{\mathrm{3}} \left(\mathrm{4}+{z}\right)^{\mathrm{4}} \right)^{\mathrm{1}/\mathrm{9}} \\ $$$$\frac{\mathrm{11}}{\mathrm{3}}\geqslant\left(\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{2}+{y}\right)^{\mathrm{3}} \left(\mathrm{4}+{z}\right)^{\mathrm{4}} \right)^{\mathrm{1}/\mathrm{9}} \\ $$$$\left(\frac{\mathrm{11}}{\mathrm{3}}\right)^{\mathrm{9}} \geqslant\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{2}+{y}\right)^{\mathrm{3}} \left(\mathrm{4}+{z}\right)^{\mathrm{4}} \\ $$
Commented by Tinkutara last updated on 18/Nov/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$