已知x、y∈R且y=x^2求证：log2[底数2](2^x+2^y)＞8/7
热心网友
只能证明log2(2^x+2^y)≥7/8,因为2^x+2^y≥2√[(2^x)×(2^y)]=2√[2^(x+x^2)],而x+x^2最小值为-1/4，所以2^x+2^y最小值为2√[2^(-1/4)]=2^(1-1/8)=2^7/8,所以log2(2^x+2^y)最小值为log2(2^7/8)=7/8,即log2(2^x+2^y)≥7/8
热心网友
∵y=x^2∴2^x+2^y=2^x+2^(x^2)=2^x+2^(2x)=2^x (1+2^x)令2^x=m(m＞0)∴2^x+2^y=(2^x)(1+2^x)=m(1+m)=m^2+m=m^2+(m/2)+(m/2)≥3[m^2*(m/2)*(m/2)]^(1/3)=3*(1/4)^(1/3)=(27/4)^(1/3)∴log2[底数2](2^x+2^y)=lg(2^x+2^y)/lg2=lg[(27/4)^(1/3)]/lg2=(1/3)(lg27-lg4)/lg2=(1/3)(3lg3-2lg2)/lg2=以lg2=0.3010,lg3=0.4771代入，上式=0.91840.9184＜8/7啊