ta có xy≥2016x+2017y ⇒1≥2016y+2017x ( do x=0;y=0 k tm) ⇒x+y≥(2016y+2017x)(x+y)=2016xy+2017yx+2016+2017 $\geq 2016+2017+2\sqrt{\frac{2016x}{y}+\frac{2017y}{x}}=(\sqrt{2016}+\sqrt{2017})^{2}$
ta có xy≥2016x+2017y ⇒1≥2016y+2017x ( do x=0;y=0 k tm) ⇒x+y≥(2016y+2017x)(x+y)=2016xy+2017yx+2016+2017 ≥2016+2017+2√2016xy2017yx =(√2016+√2017)2