已知：X+2Y=12（X＞0，Y＞0）.求证:log(2)X+log(4)Y≤7/2.
热心网友
x+2y＝(x/2)+(x/2)＋2y≥3[(三次根号下)(x/2)*(x/2)*2y] ＝3[(三次根号下)(x^2)y/2]所以 3[(三次根号下)(x^2)y/2]≤12(三次根号下)[(x^2)y/2]≤4即 (x^2)y≤128log(2)x+log(4)y＝log(4)(x^2)＋log(4)y＝log(4)[(x^2)y]≤log(4)128＝[log(2)128]/[log(2)4]＝7/2