Saudi Arabian IMO Booklet

First, we will show that $4\sqrt{b^2+c^2}\le (3-a{)}^2$. Notice that $b^2+c^2=2-a^2\ge 0$, so we need to prove $4\sqrt{2-a^2}\le (3-a{)}^2$. According to the AM-GM inequality, we have $$4\sqrt{2-a^2}=4\sqrt{1\cdot (2-a^2)}\le 4\cdot \frac{1+2-a^2}{2}=2(3-a^2).$$ We need to prove $$2(3-a^2)\le (3-a{)}^2⟺3(a^2-2a+1)\ge 0⟺3(a-1{)}^2\ge 0,$$ which is true. From this, one can conclude that, $\frac{\sqrt{b^2+c^2}}{3-a}\le \frac{3-a}{4}$. Similarly, $\frac{\sqrt{c^2+a^2}}{3-b}\le \frac{3-b}{4}$. Therefore $$P\le \frac{3-a}{4}+\frac{3-b}{4}+a+b-2022c\le \frac{6+3(a+b)}{4}.$$ We have the familiar inequality $$(a+b{)}^2\le 2(a^2+b^2)\le 2(a^2+b^2+c^2)=4$$ so $a+b\le 2$. From this it follows that $P\le \frac{6+3\cdot 2}{4}=3$. So the maximum value of $P$ of 3, is obtained when $a=b=1$, $c=0$. $\square$