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Can anyone see how to prove the following?

If $x$ and $y$ are real numbers with $y\geq 0$ and $y(y+1) \leq (x+1)^2$ then $y(y-1) \leq x^2$.

It seems it is true at least according to Mathematica.

2 Answers2

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Given $y^2 + y \le x^2 + 2x + 1$, if possible, let $x^2 < y(y-1)$. Clearly $y > 1$.

Then $x^2 + (2x + 1) < y^2 - y + (2x + 1)$
So $y^2 + y < y^2 - y + 2x + 1$, which resolves to $y < x + \frac{1}{2}$.

Hence we also have $y - 1 < x - \frac{1}{2}$.
As $y > 1$, the LHS is positive, and we can multiply the last two to get
$y(y-1) < x^2 - \frac{1}{4} \implies y(y-1) < x^2$, a contradiction.

Macavity
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  • @ShreevatsaR Thanks. Found your direct approach quite neat (+1), especially after the condensing. – Macavity Apr 16 '13 at 10:38
  • Are you sure this is correct? You had $y<x+1/2$, and in the last step you multiplied $y-1$ by $y$ and $x-1/2$ by $x+1/2$. However, this inequality is not in the same proportion anymore, because $y$ does not equal $x+1/2$. It's like if I try to prove that $a<b$, and to solve this I say $a<b+5$. While this is true, this changes the inequality. – Ovi Apr 21 '13 at 13:54
  • @Aryabhata: Oh, somehow I missed that. Sorry. I'll delete my errors, to keep the comments clear. I was wrong. – Glen O Apr 21 '13 at 14:49
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    @Ovi I seem to have missed some of the dialogue, however hope your concern has been resolved. Here we are not worried about preserving proportion, only about preserving order. In your example, $a < b \implies a < b+5$, though $a:b \neq a:(b+5)$ in general. – Macavity Apr 21 '13 at 17:55
  • I followed the proof, but I don't understand how to justify multiplying the equations (or how to think of doing that in the first place). I would appreciate any clarification! – user121955 Jul 22 '14 at 18:45
  • @user121955 To show contradiction, we need to find a good upper bound for $y(y-1)$. Hence the motive to multiply. Multiplication is justified as both LHS and RHS are non-negative. – Macavity Jul 23 '14 at 02:54
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Yes, it's true.

First note that if $0 \le y \le 1$, then $y - 1 \le 0$, so $y(y-1) \le 0 \le x^2$, and the inequality is true.

So we can assume $y > 1$. Then, as $y(y+1) = (y + \frac12)^2 - \frac14$ (by completing the square), the given $y(y+1) \le (x+1)^2$ becomes $(y + \frac12)^2 - \frac14 \le (x+1)^2$, and in particular $(y + \frac12)^2 \le (x+1)^2$ so $y \le x+\frac12$.

Then $y(y-1) = (y-\frac12)^2 - \frac14 \le x^2 - \frac14$, which is a stronger inequality than what you wanted.

If you're interested in how to arrive at this, the first version of this answer does this whole thing in a more straightforward (but longer) way; this is cleaner.

Community
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ShreevatsaR
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