来自李怡勇的问题
用均值不等式求y=2x^2+1/x+1(x>-1)的最值
用均值不等式求y=2x^2+1/x+1(x>-1)的最值
1回答
2020-05-04 19:34
用均值不等式求y=2x^2+1/x+1(x>-1)的最值
用均值不等式求y=2x^2+1/x+1(x>-1)的最值
y=(2x²+1)/(x+1)
=[2(x+1-1)²+1]/(x+1)
=[2(x+1)²-4(x+1)+3]/(x+1)
=2(x+1)+3/(x+1)-4
≥2√[2(x+1)•3/(x+1)]-4
=2√6-4
当且仅当2(x+1)=3/(x+1),即x=√6/4-1时,
y有最小值为2√6-4