Question Number 144156 by mathdanisur last updated on 22/Jun/21
Commented by mitica last updated on 22/Jun/21
$${x}=\mathrm{2},{y}=\mathrm{8}\:{false} \\ $$$$,,\leqslant,,\:{or},,\geqslant,, \\ $$
Commented by mathdanisur last updated on 22/Jun/21
$${Sorry}\:{Sir},\:``\geqslant'' \\ $$
Answered by mitica last updated on 22/Jun/21
$$\left(\frac{{x}}{{x}+{y}}\right)^{\frac{{x}+{y}}{\mathrm{2}}} \leqslant\left(\frac{{x}}{{x}+{y}}\right)^{\sqrt{{xy}}} \\ $$$$\left(\frac{{y}}{{x}+{y}}\right)^{\frac{{x}+{y}}{\mathrm{2}}} \leqslant\left(\frac{{y}}{{x}+{y}}\right)^{\sqrt{{xy}}} \\ $$
Commented by mathdanisur last updated on 22/Jun/21
$${cool}\:{thanks}\:{Sir},\:{could}\:{you}\:{simplyfy}\:{it}\:{please} \\ $$