设x.y均为实数，求证：6x/36x+1+1≤(5/6)-y+(y2/3)
热心网友
36^(x+1)+1≥2√(36^(x+1)*1)=2*6^(x+1)=12*6^x所以6^x/[36^(x+1) +1]≤6^x/(12*6^x) =1/12(5/6)-y+(y^2/3)=(y^2-3y+9/4)/3-3/4+5/6=(y-3/2)^2/3+1/12≥1/12所以 6^x/[36^(x+1) +1]≤(5/6)-y+(y^2/3)
设x.y均为实数，求证：6x/36x+1+1≤(5/6)-y+(y2/3)
36^(x+1)+1≥2√(36^(x+1)*1)=2*6^(x+1)=12*6^x所以6^x/[36^(x+1) +1]≤6^x/(12*6^x) =1/12(5/6)-y+(y^2/3)=(y^2-3y+9/4)/3-3/4+5/6=(y-3/2)^2/3+1/12≥1/12所以 6^x/[36^(x+1) +1]≤(5/6)-y+(y^2/3)