已知 $a,b,c$ 是三角形三边，证明：$\dfrac{a}{2a+c-b}+\dfrac{b}{2b+a-c}+\dfrac{c}{2c+b-a}\geqslant \dfrac 32$．
解析 设 $a=y+z$，$b=z+x$，$c=x+y$，则\[LHS=\sum_{cyc}\dfrac{y+z}{3y+z} =\sum_{cyc}\left(\dfrac 13+\dfrac 23\cdot \dfrac{z}{3y+z}\right),\]而根据柯西不等式\[\sum_{cyc}\dfrac{z}{3y+z}\geqslant \dfrac{\left(\sum_{cyc}z\right)^2}{\sum_{cyc}(3yz+z^2)}=\dfrac{\sum_{cyc}x^2+2\sum_{cyc}xy}{\sum_{cyc}x^2+3\sum_{cyc}xy}\geqslant\dfrac 34,\]因此原不等式得证．