\(a,b,c\in R^+求证：\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}\)
证明：
\(\because a,b,c\in R^+,\therefore\exists x,y,使得b=ax,c=ay\)
\(\therefore LHS=\frac{1}{x+y}+\frac{x}{1+y}+\frac{y}{1+x}\)
\(\therefore 令f(x,y)=\frac{1}{x+y}+\frac{x}{1+y}+\frac{y}{1+x}\)
令 \(\begin{cases}\begin{split}\frac{\partial f}{\partial x}&=-\frac{1}{(x+y)^2} + \frac{1}{1+y} - \frac{y}{(1+x)^2}=0\\\\\frac{\partial f}{\partial y}&=-\frac{1}{(x+y)^2} + \frac{1}{1+x} - \frac{x}{(1+y)^2}=0\end{split}\end{cases}\)
\(解得x=y=1,由题设,f(x,y)在x,y\in R^+有最小值,即f(1,1)=\frac{3}{2}(可以用二阶偏微分验证其是最小值)\)
\(\therefore \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}\)