Q:

Proving an Inequality (terms won't cancel out)

Problem: If $x$ and $y$ are real numbers such that $y \geq 0$ and $y(y+1) \leq (x+1)^2$, prove that $y(y-1) \leq x^2$.
This is what I tried:
\begin{align}
y(y+1) \leq (x+1)^2 &\implies y^2 + y \leq x^2 + 2x + 1 \\
                    &\implies x^2 \geq y^2 + y - 2x - 1 \\
                    &\implies x^2 \geq y^2 - y - 2x - 1 \tag{because $y \geq 0$}\\
                    &\implies x^2 \geq y(y-1) -2x - 1
\end{align}
I think I'm approaching this wrong, because if I start with the inequality $y(y+1) \leq (x+1)^2$, then the $2x + 1$ that results from the expansion of the RHS will never cancel out.
So, I tried this instead:
\begin{align}
y(y+1) \leq (x+1)^2 &\implies y^2 + y \leq x^2 + 2x + 1 \\
                    &\implies x^2 \geq y^2 + y - 2x - 1 \\
                    &\implies x^2 \geq y^2 - y - 2x - 1 \tag{same as before}
\end{align}
Then, I supposed that $y(y-1) \leq x^2$ is true. Consequently, because $y^2 - y - 2x - 1 \leq x^2$ (the last line), it is sufficient to show that $y(y-1) \leq y^2 - y - 2x - 1$. Again, this doesn't work because $-2x - 1$ doesn't cancel out.
Now I'm very much stuck and don't know how to proceed. If anyone can give me a hint, it'll be very much appreciated. Thanks!

A:

Consider two cases: $2x+1 \leq 2y$, and $2x+1 > 2y$. Note that we can assume $y > 1$ since if $y < 1$ the inequality is true trivially.
Specifically, if $2x+1 \leq 2y \to y(y-1) = y(y+1) - 2y \leq (x+1)^2 - 2y = x^2 + (2x+1-2y) \leq x^2 + 0 = x^2$, and if $2x+1 > 2y \to x > \dfrac{2y-1}{2} \to x^2 > \dfrac{(2y-1)^2}{4} = \dfrac{4y^2-4y+1}{4} = y(y-1) + \dfrac{1}{4} > y(y-1)$.

