Prove that $$0<\frac{\Gamma(x+y)}{\Gamma(xy)-1}\leq3$$ for all $x>0,y>0, xy>2.$ And equality holds $x=y=2.$
2026-04-08 11:21:06.1775647266
Inequality for Gamma function
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Both claims are wrong: For $x=y=2$ $$\frac{\Gamma(x+y)}{\Gamma(xy)-1}=\frac{6}{6-1}=\frac{6}{5}$$ For $x=y=1.5$ $$\frac{\Gamma(x+y)}{\Gamma(xy)-1}\approx 15.0372439$$