# Can these logarithmic inequalities show existence of a prime between (x-1)^2 and x^2

SOME PROPERTIES OF THE SERIES OF COMPOSED NUMBERS, p2 gives the bounds

$$l(x)=\frac{x}{\log(x)-28/29}<\pi(x) < u(x)=\frac{x}{\log(x)-1.12} \qquad (1)$$

for $x \geq 3299$.

$$x\pi(x) < 2 \pi(x^2)$$ for $x \geq 67$.

To show $\pi(x^2) > x\pi(x)/2 > \pi((x-1)^2)$ tried the bounds in (1) leading to:

$$xl(x) /2 > u((x-1)^2) \iff$$ $$\frac{29\, x^{2}}{2 \, {\left(29 \, \log\left(x\right) - 28\right)}} - \frac{{\left(x - 1\right)}^{2}}{\log\left({\left(x - 1\right)}^{2}\right) - 1.12} >0 \qquad (2)$$

Experimentally the LHS of (2) is increasing up to $10^{120}$ and is quite large.

1. Is this reasoning correct?
2. Can (2) be proved?

All CAS I tried failed to solve (2) though the limit at infinity is $+\infty$.

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Note that, in the second paper, the inequality $$x\pi(x)<2\pi(x^2)$$ is derived from the inequality $$xy\pi(x)\pi(y)>4\pi^2(xy)$$ by setting $x=y$. The proof for theorem 3 seems correct to me, so the sign flip is definitely a typo :)
Yes, it should be $x\pi(x)>2\pi(x^2)$. – Gjergji Zaimi May 5 '12 at 12:36