Proof that logarithmic inequality is true I have following inequality
$$ \log_p\log_p x -\frac{1}{2}\log_p(\log_p x + \log_p\log_p x+1) \geq -\log_p \sqrt{2\pi} + \log_p e^2 + \frac{1}{2}\log_p (p-1) -2$$
where d $e$ is Euler's number. We only have to consider $p \geq 2$ and $x\geq3$, but we can also increase the lower bound on $x$ if it helps. 
I want to proof, that the inequality is true. 
If it helps, we can make further conditions on $x$ and $p$.
I would appreciate, if someone can help me with this problem.
 A: I've tried myself to show that the inequality is true.
At first I take the exponent to the base $p$ on both side of the origin inequality.
$$
p^{\log_p \log_px-\frac{1}{2}\log_p(\log_px+\log_p \log_p x+1)} \geq p^{-\log_p \sqrt{2\pi}+\log_p e^2 + \frac{1}{2}\log (p-1)-2}
$$
Then I seperate the long exponents on both sides $p^{x+y}=p^xp^y$:
$$
p^{\log_p \log_px}p^{-\frac{1}{2}\log_p(\log_px+\log_p \log_p x+1)} \geq p^{-\log_p \sqrt{2\pi}}p^{\log_p e^2}p^{\frac{1}{2}\log (p-1)}p^{-2}
$$
Then I used the property of the logarithm $p^{\log_px}=x$:
$$
\frac{\log_px}{\sqrt{\log_px+\log_p \log_p x+1}} \geq \frac{1}{\sqrt{2\pi}}e^2\sqrt{p-1}\frac{1}{p^2}
$$
By now, we know for sure, that for $p\geq2$ and $x\geq2$ all terms (except the big root in the LHS-denominator) are positive and we show that also $$\log_px+\log_p \log_p x+1$$ of the long root in the LHS-denominator is positive for $p,x\geq2$:
$$\log_px+\log_p \log_p x+1 > 0$$
We exponate and seperate again:
$$
p^{\log_px}p^{\log_p \log_p x}p > p^0
$$ 
$$
xp \log_p x > 1
$$
We change to the natural logarithm:
$$
x \frac{\ln x}{\ln p} > \frac{1}{p}
$$
$$
x \ln x > \frac{\ln p}{p}
$$
LHS is monoton increasing in $x$. RHS is monoton decreasing in $p$ and upper bounded by $1$. Then I check, if the minimum possible value of the RHS (with $x=2$) is still larger than 1 and fortunately it is. 
So back to our origin inequlity, I rearrange the terms: 
$$
p^2 \log_p x \sqrt{2\pi} \geq e^2 \sqrt{(\log_p x + \log_p \log_p x +1) (p-1)}  
$$
Maybe someone see possible steps from here.
A: I fear it is possible to calculate counterexamples.
Simplifying the inequality one has
$$\frac{\log_p(x)}{\sqrt{\log_p(px\log_p(x))}}\ge(\frac ep)^2\sqrt{\frac{p-1}{2\pi}}\approx0.73695172253$$
Take $p=2$ so you have $$\frac{\log_2(x)}{\sqrt{\log_2(2x\log_2(x))}}\ge(\frac e2)^2\sqrt{\frac{2-1}{2\pi}}\approx0.73695172253$$
This is not true for $x=2.1\ge2$; in fact we have  $$\frac{\log_2(2.1)}{\sqrt{\log_2(2\cdot(2.1)\log_2(2.1))}}\approx0.726874707405\color{red}{\lt}0.73695172253$$
Actually the inequality fails for $p=2$ and (approximately) $1.396\lt x\lt2.152$
and because of the given condition $x\ge2$ we have counterexamples in all the semi-open interval $[2,\space2.152[$.
