Let $u(x,y), x^2+y^2 \leq 1$, a solution of
$$u_{xx}(x,y)+2u_{yy}(x,y)+e^{u(x,y)}=0, x^2+y^2\leq 1$$
Show that $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y) $.
We suppose that $\min_{x^2+y^2 \leq 1} u(x,y) \neq \min_{x^2+y^2=1} u(x,y) $.
At the solution it is said that since $\{ (x,y) | x^2+y^2 \leq 1\} \supseteq \{ (x,y)| x^2+y^2 =1 \}$ it has to hold that $\min_{x^2+y^2 \leq 1} u(x,y)<\min_{x^2+y^2=1} u(x,y)$ .
I haven't understood how we deduce that the above inequality holds. Could you explain it to me?
If $A\subset B$ and $f\colon A\to\mathbb{R}$, then $$ \inf_{x\in A}f(x)\le\inf_{x\in B}f(x). $$ Strict inequality may not hold.
The inequality $\min_{x^2+y^2 \leq 1} u(x,y)<\min_{x^2+y^2=1} u(x,y)$ must be incorrect, since you are asked to prove that $\min_{x^2+y^2 \leq 1} u(x,y)=\min_{x^2+y^2=1} u(x,y)$.