等价于
新问题已知 $x,y$ 是正实数，且 $\dfrac{1}{x+1}+\dfrac{1}{y+1}=\dfrac 13$，求 $x^2+y^2-13xy$ 的最小值．
解设 $\dfrac{1}{x+1}=a$，$\dfrac{1}{y+1}=b$，则\[a+b=\dfrac 13,\]且\[\begin{split} x^2+y^2-13xy&=\left(\dfrac 1a-1\right)^2+\left(\dfrac 1b-1\right)^2-13\left(\dfrac 1a-1\right)\left(\dfrac 1b-1\right)\\
&=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{11}{a}+\dfrac{11}{b}-\dfrac{13}{ab}-11\\
&=\dfrac{a^2+b^2+11(a+b)ab-13ab}{a^2b^2}-11\\
&=\dfrac{\dfrac 19-2ab+\dfrac{11}3ab-13ab}{a^2b^2}-11\\
&=\dfrac{1}{9a^2b^2}-\dfrac {34}{3ab}-11\\
&\geqslant -300,\end{split}\]等号当 $\dfrac{1}{ab}=51$ 时取得，因此所求的最小值为 $-300$．