<s>[INST] Max and Min of $f(x,y)$ Let $f(x,y)=x(y \log y-y)-y \log x$. Find $\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(x,y)})$. 
 [/INST] Note that $$\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(x,y)}) \leq \smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}f(x,\frac{1}{2})=\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(-\frac{x\log 2+x+\log x}{2})$$
Since $-\frac{x\log 2+x+\log x}{2}$ is a decreasing function of $x$, we have 
$$\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(-\frac{x\log 2+x+\log x}{2})=-\frac{\frac{1}{2}\log 2+\frac{1}{2}+\log \frac{1}{2}}{2}=\frac{\log 2-1}{4}$$
with equality when $x=\frac{1}{2}$. We claim that equality holds throughout and thus
$$\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(x,y)})=\frac{\log 2-1}{4}$$
It now suffices to show that 
$$\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(\frac{1}{2},y)}=f(\frac{1}{2}, \frac{1}{2})$$
We have 
$$f(\frac{1}{2}, y)=\frac{1}{2}y\log y-\frac{1}{2}y+y\log 2$$
$$\frac{d f(\frac{1}{2}, y)}{dy}=\frac{1}{2}\log y+\log 2>\frac{1}{2}\log \frac{1}{2}+\log 2=\frac{1}{2}\log 2>0$$
Therefore $f(\frac{1}{2}, y)$ is a non-decreasing function of $y$, and we are done.
</s>