x3+y3=x-y

12. 已知 $x>0$，$y>0$，且 $x^{3}+y^{3}=x-y$，则（　　）

A. $x+y\geq\sqrt{\dfrac{2\sqrt{3}}{3}}-1$　　B. $x+y\leq\sqrt{\dfrac{2\sqrt{3}}{3}}$　　C. $x^{2}+y^{2}<1$　　D. $x^{2}+y^{2}>\dfrac{1}{2}$

思路一：

解： 对 A,B 选项

设 $x+y=t$，则 $x=t-y$

$\therefore\left(t-y\right)^{3}+y^{3}=t-y-y$

$\therefore t^{3}-3t^{2}y+3ty^{2}-y^{3}+y^{3}=t-2y$

$\therefore 3ty^{2}+\left(2-3t^{2}\right)y+t^{3}-t=0$

$\therefore\Delta=\left(2-3t^{2}\right)^{2}-12t\left(t^{3}-t\right)\geq0$

$\therefore t^{4}\leq\dfrac{4}{3}$

$\therefore t\leq\sqrt{\dfrac{2\sqrt{3}}{3}}$，故 B 正确

对 C,D 选项

设 $x^{2}+y^{2}=r^{2}$，$x=r\sin\theta$，$y=r\cos\theta$，$\theta\in\left(0,\dfrac{\pi}{2}\right)$

$\therefore r^{3}\sin^{3}\theta+r^{3}\cos^{3}\theta=r\sin\theta-r\cos\theta$

$\therefore r^{2}=\dfrac{\sin\theta-\cos\theta}{\sin^{3}\theta+\cos^{3}\theta}$

设 $f(\theta)=\dfrac{\sin\theta-\cos\theta}{\sin^{3}\theta+\cos^{3}\theta}$

则 $f'(\theta)=\dfrac{\left(\cos\theta+\sin\theta\right)\left(\sin^{3}\theta+\cos^{3}\theta\right)-\left(\sin\theta-\cos\theta\right)\left(3\sin^{2}\theta\cos\theta-3\cos^{2}\theta\sin\theta\right)}{\left(\sin^{3}\theta+\cos^{3}\theta\right)^{2}}$

$=\dfrac{\left(\cos\theta+\sin\theta\right)^{2}\left(1-\sin\theta\cos\theta\right)-\left(\sin\theta-\cos\theta\right)^{2}\cdot3\sin\theta\cos\theta}{\left(\sin^{3}\theta+\cos^{3}\theta\right)^{2}}$

$=\dfrac{\left(1+2\sin\theta\cos\theta\right)\left(1-\sin\theta\cos\theta\right)-\left(1-2\sin\theta\cos\theta\right)\cdot3\sin\theta\cos\theta}{\left(\sin^{3}\theta+\cos^{3}\theta\right)^{2}}$

$=\dfrac{4\sin^{2}\theta\cos^{2}\theta-2\sin\theta\cos\theta+1}{\left(\sin^{3}\theta+\cos^{3}\theta\right)^{2}}=\dfrac{\left(\sin2\theta-\dfrac{1}{2}\right)^{2}+\dfrac{3}{4}}{\left(\sin^{3}\theta+\cos^{3}\theta\right)^{2}}>0$

$\therefore f(\theta)$ 在 $\left(0,\dfrac{\pi}{2}\right)$ 上单增

$\therefore x^{2}+y^{2}=r^{2}<f\left(\dfrac{\pi}{2}\right)=1$，故 C 正确

思路二：

解析： 因为 $x^{3}+y^{3}=x-y$，所以 $\left(x+y\right)\left(x^{2}-xy+y^{2}\right)=x-y$，$x+y=\dfrac{x-y}{x^{2}-xy+y^{2}}$

所以 $\left(x+y\right)^{2}=\dfrac{x^{2}-y^{2}}{x^{2}-xy+y^{2}}=\dfrac{\left(\dfrac{x}{y}\right)^{2}-1}{\left(\dfrac{x}{y}\right)^{2}-\dfrac{x}{y}+1}$

令 $\dfrac{x}{y}=t$，因为 $x^{3}+y^{3}=x-y$，$x,y>0$，所以 $x-y>0$，即 $t=\dfrac{x}{y}>1$

$\left(x+y\right)^{2}=\dfrac{t^{2}-1}{t^{2}-t+1}=1+\dfrac{t-2}{t^{2}-t+1}$，当 $t=2$ 时，$\left(x+y\right)^{2}=1$

当 $t>1$ 且 $t\neq2$ 时，令 $u=t-2$，则 $\left(x+y\right)^{2}=1+\dfrac{t-2}{t^{2}-t+1}=1+\dfrac{1}{u+\dfrac{3}{u}+3}$

因为 $u\in(-1,0)\cup(0,+\infty)$，所以 $\left(x+y\right)^{2}=1+\dfrac{1}{u+\dfrac{3}{u}+3}\in(0,1)\cup\left(1,\dfrac{2\sqrt{3}}{3}\right]$

所以 $0<\left(x+y\right)^{2}\leq\dfrac{2\sqrt{3}}{3}$，$x+y\leq\sqrt{\dfrac{2\sqrt{3}}{3}}$

因为 $y<x$，所以当 $x\to0$ 时，$x+y<2x\to0$，A、D 错误

因为 $x^{3}+y^{3}=x-y$，所以 $y^{3}+y+x^{3}-x=0$

令 $f(t)=t^{3}+t+x^{3}-x$，$f(y)=0$，

因为 $f(t)$ 在 $(0,+\infty)$ 上单调递增，$f(t)$ 的零点 $y$ 满足 $y>0$

所以 $f(0)=x^{3}-x<0$，解得 $x<1$

所以要证 $x^{2}+y^{2}<1$，即证 $y<\sqrt{1-x^{2}}$

因为 $f(t)$ 在 $(0,+\infty)$ 上单调递增，所以即证 $f\left(\sqrt{1-x^{2}}\right)>0$

因为 $f\left(\sqrt{1-x^{2}}\right)=\left(\sqrt{1-x^{2}}\right)^{3}+\sqrt{1-x^{2}}+x^{3}-x=\sqrt{1-x^{2}}\left(\dfrac{x+2\left(\sqrt{1-x^{2}}\right)^{3}}{\sqrt{1-x^{2}}+x}\right)>0$

所以 $f\left(\sqrt{1-x^{2}}\right)>0$ 成立，即 $x^{2}+y^{2}<1$ 成立

故选 BC。