Question
Prove or disprove: if $x$ and $y$ are real numbers with $y \geq 0$ and $y(y+1) \leq(x+1)^{2},$ then $y(y-1) \leq x^{2}$
Step 1
Then, we can divide the inequality $y(y+1) \leq(x+1)^{2}$ by $y$ to get $y+1 \leq \frac{(x+1)^{2}}{y}$. Show more…
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