# No positive real such that $\left\lfloor\frac{25}{x}+\frac{49}{a}\right\rfloor=\left\lfloor\frac{144}{x+a}-1\right\rfloor$

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

$$\left\lfloor\frac{25}{x}+\frac{49}{a}\right\rfloor=\left\lfloor\frac{144}{x+a}-1\right\rfloor$$

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?

• use: '\lfloor' & '\rfloor'.. Commented Feb 13, 2020 at 13:14
• The expression seems like one has to use CS inequality or more specifically, Titu's Lemma Commented Feb 13, 2020 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.
• 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.
• Jam, $\frac{5^2}{x}+\frac{7^2}{a} \geq \frac{(5+7)^2}{x+a}$ from Cauchy-Schwarz :)