pggp 26.10.2020 16:52 Let $x$, $y$ be real numbers, such that $x^2 + x \leq y$. Prove that $y^2 + y \geq x$.
Faustus 26.10.2020 18:07 $y^2+y\ge (x^2+x)^2+(x^2+x)= ((x^2+x)^2+x^2)+x\ge x$ since squares of reals are greater than zero.