已知\(k>0,b>0\),且\(\forall x>-4,kx+b\geqslant {\ln}(x+4)\),则\(\dfrac{b}{k}\)的最小值为\(\underline{\qquad\qquad}.\)
解析:
将原题条件重新叙述\(:\)
\[\forall x>0,k(x-4)+b-{\ln}x\geqslant 0,k,b>0.\]
记上述不等式左侧为\(f(x)\)则有\[ \forall x>0,f(x)\geqslant f\left(\dfrac{1}{k}\right)=b+1-4k+{\ln}k\geqslant 0.\]
所以\(b\geqslant 4k-1-{\ln}k\),从而\[ \dfrac{b}{k}\geqslant 4-\dfrac{1+{\ln}k}{k}\geqslant 3.\]因此当且仅当\((b,k)=(3,1)\)时,\(\dfrac{b}{k}\)取得最小值\(3\).