Let $a>0$. Prove $\nexists x\in\mathbb R^+$ s.t.


I know that $$k\in\mathbb Z\implies\left(\forall x\in\mathbb R\right) \lfloor x+k\rfloor=\lfloor x\rfloor + k.$$

I can say that $\left\lfloor\frac{144}{x+a}-1\right\rfloor=\left\lfloor\frac{144}{x+a}\right\rfloor-1$.

How should I proceed after this? Does this help in any way?

  • $\begingroup$ use: '\lfloor' & '\rfloor'.. $\endgroup$ – Invisible Feb 13 at 13:14
  • $\begingroup$ The expression seems like one has to use CS inequality or more specifically, Titu's Lemma $\endgroup$ – user35508 Feb 13 at 13:21

Hint: The floor function, $\lfloor x\rfloor$, is bounded by $x-1< \lfloor x\rfloor\le x$ so you have $\text{RHS}\le \left(\frac{144}{x+a}-1\right)$ and $\left(\frac{25}{x}+\frac{49}{a}\right)-1<\text{LHS}$. By considering their (single) intersection, consider how you can form an inequality with $\left(\frac{144}{x+a}-1\right)$ and $\left(\frac{25}{x}+\frac{49}{a}\right)-1$, then check whether this intersection is a solution and find an inequality relating all four terms.

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  • $\begingroup$ A good thing to notice is that $25,49,144$ are all perfect squares so there might appear to be a hidden quadratic equation in the problem. $\endgroup$ – Jam Feb 13 at 13:58
  • 1
    $\begingroup$ Jam, $\frac{5^2}{x}+\frac{7^2}{a} \geq \frac{(5+7)^2}{x+a}$ from Cauchy-Schwarz :) $\endgroup$ – LHF Feb 13 at 14:08
  • $\begingroup$ @Atticus You can use CS but I think it's just as easy to prove it without it in this case. $\endgroup$ – Jam Feb 13 at 14:13

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