Preselection tests for the full-time training

First solution. Because $-1\le x,y\le 1$, we have $(1-x^2),(1-y^2)\ge 0$. Therefore, by applying AM-GM, we obtain $$2\sqrt{(1-x^2)(1-y^2)}\le (1-x^2)+(1-y^2).$$ So, it remains to prove that $$(1-x^2)+(1-y^2)\le 2(1-x)(1-y)+1$$ which is equivalent to $$x^2+y^2-2x-2y+2xy+1\ge 0.$$ But $$x^2+y^2-2x-2y+2xy+1=(x+y-1{)}^2$$ This ends the proof. The equality holds when $1-x^2=1-y^2$ and $x+y-1=0$. This is equivalent to $x=y=\frac{1}{2}$.

Second solution. Because $-1\le x,y\le 1$, there exist $-\frac{\pi }{2}\le \theta ,\varphi \le \frac{\pi }{2}$ such that $x=\sin \theta$ and $y=\sin \varphi$. The inequality to prove becomes $$2\cos \theta \cos \varphi \le 2(1-\sin \theta )(1-\sin \varphi )+1,$$ or, equivalently $$2[\cos (\theta +\varphi )+\sin \theta +\sin \varphi ]\le 3.$$ Using the relations $\cos (\theta +\varphi )=1-2{\sin }^2\frac{\theta +\varphi }{2}$ and $\sin \theta +\sin \varphi =2\sin \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}$, the inequality simplifies to $${\cos }^2\frac{\theta -\varphi }{2}\le 1+{\left(2\sin \frac{\theta +\varphi }{2}-\cos \frac{\theta -\varphi }{2}\right)}^2,$$ which is satisfied. The equality holds when $\theta =\varphi =\frac{\pi }{6}$, which means $x=y=\frac{1}{2}$.