三元代数式最值探索

已知$x,y,z\geqslant 0$，$p$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$$\lambda _{1}f(x_1)+\lambda _{2}f(x_2)+\cdots+\lambda _{n}f(x_n)\geqslant f\left(\lambda _{1}x_1+\lambda _{2}x_2+\cdots+\lambda _{n}x_n\right).$$