已知x,y,z⩾,p是一个给定的实数,求f_p=\displaystyle \sum\limits_{cyc}\left(\dfrac{x}{y+z}\right)^p的最小值关于p的表达式S(p).
分析与解 不妨设x+y+z=1且x\geqslant y\geqslant z,则z\in\left[0,\dfrac 13\right],记r={\log_2}3-1.
情形一 p\leqslant 0.
此时\begin{split} f_p&=\left(\dfrac{y+z}x\right)^{-p}+\left(\dfrac{z+x}y\right)^{-p}+\left(\dfrac{x+y}z\right)^{-p}\\ &\geqslant 3\left(\dfrac{y+z}x\cdot \dfrac{z+x}y\cdot \dfrac{x+y}z\right)^{-\frac p3}\\ &\geqslant 3\left(\dfrac{2\sqrt{yz}}x\cdot \dfrac{2\sqrt{xy}}y\cdot \dfrac{2\sqrt{yz}}x\right)^{-\frac p3}\\ &=\dfrac{3}{2^p},\end{split} 等号当且仅当x=y=z时取得.因此f_p的最小值S(p)=\dfrac{3}{2^p}.
情形二 0<p\leqslant \dfrac 12.
此时\left(\dfrac{x}{y+z}\right)^p=\dfrac{x^{2p}}{x^p(y+z)^p}\geqslant \dfrac{2x^{2p}}{x^{2p}+(y+z)^{2p}}\geqslant \dfrac{2x^{2p}}{x^{2p}+y^{2p}+z^{2p}},因此有f_p\geqslant \sum_{cyc}\dfrac{2x^{2p}}{x^{2p}+y^{2p}+z^{2p}}=2,等号当x=y=\dfrac 12,z=0时取得.因此f_p的最小值S(p)=2.
情形三 \dfrac 12<p\leqslant r.
此时根据幂平均不等式和琴生不等式(注意:因为函数y=\sqrt{\dfrac{x}{1-x}}的两阶导函数为y''=\dfrac 14x^{-\frac 32}(1-x)^{-\frac 52}(4x-1),所以它在区间\left[\dfrac 14,1\right]上为下凸函数),有\begin{split} f_p&\geqslant 2\left[\dfrac 12\left(\sqrt{\dfrac{x}{1-x}}+\sqrt{\dfrac{y}{1-y}}\right)\right]^{2p}+\left(\dfrac{z}{1-z}\right)^p\\ &\geqslant 2\left(\sqrt{\dfrac{\frac{x+y}2}{1-\frac{x+y}2}}\right)^{2p}+\left(\dfrac{z}{1-z}\right)^p\\ &=2\left[\dfrac{x+y}{2-(x+y)}\right]^p+\left({\dfrac{z}{1-z}}\right)^p\\ &=2\left(\dfrac{1-z}{1+z}\right)^p+\left(\dfrac{z}{1-z}\right)^p,\end{split} 令t=\dfrac{z}{1-z},则t\in \left[0,\dfrac 12\right],且z=\dfrac{t}{1+t},上式变为\varphi(t)=2(1+2t)^{-p}+t^p,t\in \left[0,\dfrac 12\right],其导函数\varphi'(t)=\dfrac{p}{t^{1-p}(1+2t)^{1+p}}\cdot \left[(1+2t)^{1+p}-4t^{1-p}\right],设\mu(t)=(1+2t)^{1+p}-4t^{1-p},则其导函数\mu'(t)=2(1+p)(1+2t)^p-4(1-p)t^{-p}.由于\mu'(t)单调递增,而\mu'\left(\dfrac 12\right)=3^p(6p-2)>0,于是\mu(t)先单调递减,再单调递增.而\mu(0)=1,\mu\left(\dfrac 12\right)=0,因此\varphi(t)先单调递增,再单调递减.此时S(p)=\min\left\{\varphi(0),\varphi\left(\dfrac 12\right)\right\}=\min\left\{\dfrac{3}{2^p},2\right\}=2.当\dfrac 12<p<r时,x=y=\dfrac 12,z=0时取到最小值;
当p=r时,x=y=\dfrac 12,z=0以及x=y=z时,同时取到最小值.
情形四 p>r.
此时根据幂平均不等式有f_p\geqslant 3\left[\dfrac 13\sum_{cyc}\left(\dfrac{x}{y+z}\right)^r\right]^{\frac pr}\geqslant 3\left(\dfrac 23\right)^{\frac pr}=3\left(\dfrac {2}{2^{r+1}}\right)^{\frac pr}=\dfrac{3}{2^p},等号当x=y=z时取得.因此S(p)=\dfrac{3}{2^p}.
综上所述,所求f_p的最小值的表达式S(p)=\begin{cases} \dfrac{3}{2^p},&p\in (-\infty,0]\cup\left({\log_2}3-1,+\infty\right),\\ 2,&p\in \left(0,{\log_2}3-1\right].\end{cases}
注 幂平均不等式与琴生不等式:
幂平均不等式 \left(\dfrac {a_1^\alpha +a_2^\alpha +\cdots+a_n^\alpha}{n}\right)^{\frac 1\alpha }\geqslant \left(\dfrac {a_1^\beta+a_2^\beta+\cdots+a_n^\beta}{n}\right)^{\frac 1\beta},其中a_i>0,\alpha >\beta.
当\alpha=2,\beta=1,n=2时,就是不等式\sqrt{\dfrac {a_1^2+a_2^2}{2}}\geqslant \dfrac {a_1+a_2}{2},当\alpha =1,\beta=-1,n=2,就是不等式\dfrac {a_1+a_2}{2}\geqslant \dfrac {2}{\dfrac 1{a_1}+\dfrac 1{a_2}}.
琴生不等式 如果f(x)是下凸函数,即对定义域内任意的x_{1},x_{2},有\dfrac 12[f(x_{1})+f(x_{2})]\geqslant f\left(\dfrac {x_{1}+x_{2}}{2}\right).那么对任意的\lambda _{i}\geqslant 0,\sum\limits_{i=1}^{n}{\lambda _{i}}=1,i=1,2,\cdots,n,有\lambda _{1}f(x_1)+\lambda _{2}f(x_2)+\cdots+\lambda _{n}f(x_n)\geqslant f\left(\lambda _{1}x_1+\lambda _{2}x_2+\cdots+\lambda _{n}x_n\right).
分类的依据是怎样找到的?