2022/003) Prove $x^2+y^2+5>xy+x+3y$

 We have

$(x-y)^2 >=0$ or $x^2 + y^2 >=2xy$

$(x-1)^2 >=0$ or $x^2 + 1 >= 2x$

$(y-3)^2 >= 0$ or $y^2 + 9 >= 6y$

Adding we get $2(x^2+y^2 + 5) >= 2(xy+x + 3y)$

Or $x^2+y^2 + 5 >= xy + x + 3y$

This is equal when x = y, y = 3, x= 1 which canot be true so  $x^2+y^2 + 5 > xy + x + 3y$