Question Number 195404 by universe last updated on 01/Aug/23
Answered by AST last updated on 01/Aug/23
$$\left(\frac{\mathrm{127}^{\mathrm{2}} }{{xy}}+\frac{{x}}{{y}}+\frac{{y}}{{x}}\right)\left(\mathrm{127}+{x}+{y}\right) \\ $$$$=\frac{\mathrm{127}^{\mathrm{3}} }{{xy}}+\frac{\mathrm{127}^{\mathrm{2}} }{{y}}+\frac{\mathrm{127}^{\mathrm{2}} }{{x}}+\frac{\mathrm{127}{x}}{{y}}+\frac{{x}^{\mathrm{2}} }{{y}}+{x}+\frac{\mathrm{127}{y}}{{x}}+{y}+\frac{{y}^{\mathrm{2}} }{{x}} \\ $$$$=\left(\frac{\mathrm{127}^{\mathrm{3}} }{{xy}}+{x}+{y}\right)+\frac{\mathrm{127}^{\mathrm{2}} }{{y}}+\frac{{y}^{\mathrm{2}} }{{x}}+\frac{\mathrm{127}{x}}{{y}}+\frac{\mathrm{127}^{\mathrm{2}} }{{x}}+\frac{{x}^{\mathrm{2}} }{{y}}+\frac{\mathrm{127}{y}}{{x}} \\ $$$$\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{127}^{\mathrm{3}} }+\mathrm{6}\sqrt[{\mathrm{6}}]{\mathrm{127}^{\mathrm{6}} }=\mathrm{9}\left(\mathrm{127}\right)=\mathrm{1143}\: \\ $$$$\left({Equality}\:{holds}\:{when}\:{x}={y}=\mathrm{127}\right) \\ $$