Croatian Junior Mathematical Olympiad

Transforming the inequality $(a+b+1{)}^2\ge 0$ yields $a^2+b^2+1+2ab+2a+2b\ge 0$, from which it follows that $2(ab+a+b)\ge -(a^2+b^2)-1$, i.e. $ab+a+b\ge -13$. The equality is attained if and only if $a+b+1=0$, i.e. $b=-a-1$. Plugging this into $a^2+b^2=25$ gives us $$\begin{array}{rl}{a}^2+(-a-1{)}^2& =25,\\ 2a^2+2a+1& =25,\\ a^2+a-12& =0,\\ (a+4)(a-3)& =0,\end{array}$$ hence we get and verify two symmetric solutions: $(a,b)=(-4,3)$ and $(a,b)=(3,-4)$.