f(x/m)-4m^2*f(x)≤f(x-1)+4f(m)
化简得,x^2/m^2 -1 -4m^2 x^2 ≤ x^2-2x -4
(1/m^2-4m^2-1)x^2≤-2x-3
1/m^2-4m^2-1≤(-2x-3)/x^2 恒成立
令y=(-2x-3)/x^2= -2(1/x)-(3/x^2)
∵x≥2/3
∴1/x ≥3/2
因为y在[3/2,+无穷)上单调增
所以 在x=3/2 取得最小值 -8/3
所以 1/m^2-4m^2-1≤-8/3
12m^4-5m^2-3≥0
(4m^2-3)(3m^2+1)≥0
所以4m^2-3≥0
答案就出来了
f(x)=x^2-1,f(x/m) - 4m^2 *f(x) ≤ f(x-1) + 4 f(m)
即(x/m)^2-1-4m^2(x^2-1)≤(x-1)^2-1+4(m^2-1),
∴x^2/m^2-4m^2x^2+4m^2≤x^2-2x+1+4m^2-4,
整理得(1/m^2-4m^2-1)x^2+2x+3≤0对3/2≤x恒成立,(改题了)
分离参数得 1/m^2-4m^2-1≤-(2/x+3/x^2)对3/2≤x恒成立,①
0<1/x≤2/3,
∴2/x+3/x^2≤8/3,于是①变为
1/m^2-4m^2-1≤-8/3,
整理得(12m^4-5m^2-3)/m^2≥0,m^2>0,
(4m^2-3)(3m^2+1)≥0,3m^2+1>0,
∴m^2≥3/4,
∴m≥√3/2或m≤-√3/2.