f(x+t)≤3xx^2+(2t-1)x+t^2+2t≤0g(x)=x^2+(2t-1)x+t^2+2t在x∈[1,m]恒≤0{g(1)=t^2+4t≤0→-4≤t≤0 g(m)=t^2+(2m+2)t+m^2-m=(t+m+1)^2-3m-1≤0→-√(3m+1)-m-1≤t≤√(3m+1)-m-1t存在∴-√(3m+1)-m-1≤0 √(3m...