Prove the second Hardy–Littlewood conjecture is true for $e^{0.112}-1<\frac{x}{y}<\frac{1}{e^{0.112}}$

Question

Prove the second Hardy-Littlewood conjecture: $$\pi\left(x+y\right)-\pi\left(y\right)\leq\pi\left(x\right)$$ is true for any integers $x,y>2$ where: $$e^{0.112}-1<\frac{x}{y}<\frac{1}{e^{0.112}}$$

Answer 1

From this theorem, we know that:

$$\frac{\pi\left(x\right)}{x}>\frac{\pi\left(y\right)-\pi\left(x\right)}{y-x}\tag{1}$$

Therefore:

$$\frac{\pi\left(x\right)}{x}>\frac{\pi\left(y\right)}{y}\tag{2}$$

Using the theorem again we also know that:

$$\frac{\pi\left(y\right)}{y}>\frac{\pi\left(x+y\right)-\pi\left(y\right)}{x+y-y}=\frac{\pi\left(x+y\right)-\pi\left(y\right)}{x}\tag{3}$$

From $\left(2\right)$ and $\left(3\right)$, we get:

$$\frac{\pi\left(x+y\right)-\pi\left(y\right)}{x}<\frac{\pi\left(x\right)}{x}$$

And finally:

$$\pi\left(x+y\right)-\pi\left(y\right)<\pi\left(x\right)$$

Now because of $\left(1\right)$ and $\left(3\right)$, we need to add those restrictions:

$$\frac{y}{x}>e^{0.112}$$

$$\frac{x+y}{y}>e^{0.112}$$

We can rearrange like this:

$$\frac{x}{y}<\frac{1}{e^{0.112}}$$ $$\frac{x}{y}>e^{0.112}-1$$

Therefore:

$$e^{0.112}-1<\frac{x}{y}<\frac{1}{e^{0.112}}$$

Note that the $x,y>5393$ from the theorem was ignored because $x,y<5393$ can be easily verified manually.