Real numbers \(x,y\) are such that \(x^2 +x \le y\). Show that \(y^2 +y \ge x\).
Today we will solve some problems using algebraic tricks, mostly
related to turning a sum into a product or using an identity involving
squares.
The ones particularly useful are: \((a+b)^2 =
a^2 +b^2 +2ab\), \((a-b)^2 = a^2 +b^2
-2ab\) and \((a-b) \times (a+b) = a^2
-b^2\). While we are at squares, it is also worth noting that any
square of a real number is never a negative number.
The evil warlock found a mathematics exercise book and replaced all the decimal numbers with the letters of the alphabet. The elves in his kingdom only know that different letters correspond to different digits \(\{0,1,2,3,4,5,6,7,8,9\}\) and the same letters correspond to the same digits. Help the elves to restore the exercise book to study.