Saudi Arabia booklet 2024

Let $P=x^2+z^2+xy+yz+zx$ then we have $$25=(2x+y+2z{)}^2=4x^2+y^2+4z^2+4xy+4yz+4zx=4P+y^2+4zx.$$ Note that $y^2+4zx\ge 0$ so $4P\le 25$ which implies that $P\le \frac{25}{4}$. The equality case is $y=z=0$ and $x=\frac{5}{2}$.

Continue, note that $(y-x)(y-z)\le 0$ so $y^2+zx\le xy+yz$. Thus $$P\ge x^2+z^2+y^2+2zx=(x+z{)}^2+y^2.$$ Using Cauchy-Schwarz inequality, one can get $$(2^2+1^2)((x+z{)}^2+y^2)\ge (2x+2z+y{)}^2=25.$$ Thus $(x+z{)}^2+y^2\ge \frac{25}{5}=5$ which implies that $P\ge 5$. The equality case is $x=y=z=1$. $\square$