Partial Proof of Second Hardy-Littlewood Conjecture (modified)? I have obtained the following partial proof of the Second Hardy-Littlewood Conjecture of which I can't find out the logical flaw in the proof. 

Problem
Prove that for all sufficiently large $x$ and $y$ such that $x\neq y$ and $x$, $y$ $\in \mathbb{R^+}$ we have,
  $$\pi(x)+\pi(y)\geq\pi(x+y)$$
Proof
For a proof of this result we use the following inequality,
  $$\dfrac{x}{\ln x-(1-\epsilon)}<\pi(x)<\dfrac{x}{\ln x-(1+\epsilon)}$$
each of which hold for all sufficiently large $x$ and for any $\epsilon>0$.
To prove,
  $$\pi(x)+\pi(y)\geq\pi(x+y)$$
Now note that, $$\pi(x)+\pi(y)>\dfrac{x}{\ln x-(1-\epsilon)}+\dfrac{y}{\ln y-(1-\epsilon)}$$ and for the same $\epsilon$ we may write, $$\pi(x+y)<\dfrac{x+y}{\ln (x+y)-(1+\epsilon)}$$
Hence,
$\pi(x)+\pi(y)\geq\pi(x+y) \\\impliedby \dfrac{x}{\ln x-(1-\epsilon)}+\dfrac{y}{\ln y-(1-\epsilon)}\geq\dfrac{x+y}{\ln (x+y)-(1+\epsilon)}\\\impliedby \left(\dfrac{x}{\ln x-(1-\epsilon)}\right)\left(\ln (x+y)-\ln x-2\epsilon\right)\geq \left(\dfrac{y}{\ln y-(1-\epsilon)}\right)\left(\ln y-\ln (x+y)+2\epsilon\right) \\\impliedby \left(\dfrac{x}{\ln x-(1-\epsilon)}\right)\geq\left(\dfrac{y}{\ln y-(1-\epsilon)}\right) \quad,\quad 2\ln \left(x+y\right)\geq\ln \left(xy\right)+4\epsilon$ 
Both of which are true. The first one for sufficiently large $x$ and $y$ (because the function $f(t)=\left(\dfrac{t}{\ln t-(1-\epsilon)}\right)$ is ultimately increasing for all $1\geq\epsilon>0$) and the second one for all $\epsilon$ satisfying the bound $0<\epsilon\leq{\ln \sqrt[4]{2}}$. Thus, $$\pi(x)+\pi(y)>\pi(x+y)$$ holds for all $x,y$ satisfying the bound for the $\epsilon$ just deduced.

What is the flaw in the proof? Can anyone help me?

Edit:-
The deleted part of the question is now asked here.
 A: It's not true that $\pi(x) + \pi(y) > 2 \pi((x+y)/2)$ for large $x$ and $y$.  
Consider $x = p-1$, $y= p+1$ where $p$ is a large prime.  Then $\pi((x+y)/2) = \pi(y) = 1 + \pi(x)$ 

Edit: The first mistake is
$$ 2\pi\left(\dfrac{x+y}{2}\right)<\dfrac{x+y}{\ln \left(\dfrac{x+y}{2}\right)}$$
All you know is 
$$ \pi\left(\dfrac{x+y}{2}\right)<\dfrac{x+y}{\ln \left(\dfrac{x+y}{2}\right)}$$
A: I noticed above that the naive approach to proving the topmost inequality without $\pi$'s fails when $y/x$ is roughly the size of $\epsilon$ or even smaller. Now, it turns out that this inequality is indeed false when $y/x$ becomes small enough in terms of $\epsilon$. How small exactly is small enough however is considerably more subtle than I thought initially, see the end of this post. But first is a counterexample with actual numbers. Plugging the same numbers into the inequalities further down should tell us which of the implications is wrong.
$\epsilon = 1/6$ seems to be the nicest epsilon that is allowed. We do not know how large $x$ and $y$ must be to satisfy De la Vallée-Poussin's inequality but it is safe to assume that they are big enough to have positive denominators in the inequality we want to disprove. So, take $y$ big enough and take $x = 11y$. (So $y/x$ is quite a bit smaller than $\epsilon$ in this example.) We plug this into the inequality and simplify notation by setting $\ln y = z$ and hence $\ln(x) = z + \ln 11$ and $\ln(x+y) = z + \ln(12)$.
Divide both sides of the inequality by $y$ and multiply with the product of the three denominators: $(z - 5/6)(z + \ln 11 -  5/6)(z + \ln 12 - 7/6)$. What we end up with is an inequality between two quadratic expressions in $z$. 
It is easy to see that both have quadratic term $12z^2$ so we can ignore that. We can also ignore the constant terms, since we are looking at 'sufficiently large $z$'.
So, just comparing the multiples of  $z$ on both sides we end up with an inequality that is equivalent to $12 \ln (12) - 11 \ln (11) \geq 24/6 = 4$ which is false.
The somewhat more general form of this equation (taking $x = ky$ instead of $x = 11y$) reads $(k+1)\ln(k+1) - k\ln(k) \geq 1/3(k + 1)$ or even more generally: 
$(k + 1)\ln(k+1) - k\ln(k) \geq 2\epsilon (k + 1).$
Keeping $\epsilon$ fixed and viewing both sides as a function of $k$ we see on the left hand side a function that in the beginning increases quite fast, but as $k$ gets bigger increases more and more and more slowly, while at the right hand side we see a function that increases slowly, but keeps this steady pace of increasing and hence will eventually (for big enough $k$) become bigger. This is the reason we can construct counterexamples to the inequality as the one above. (Provided I got it finally right this time.)
On the other hand it seems that when we keep $k$ fixed we can choose $\epsilon$ so small that the inequality does hold. Hence the situation seems to be as follows: keeping $y$ fixed and increasing $x$ we might from time to time reach a  counterexample to the Hardy-Littlewood inequality. However keeping $k = x/y$ fixed and moving $y$ (and hence $x$) up it seems that we will eventually reach a point above which the weaker inequality $\pi((k+1)y) \leq \pi(ky) + \pi(y)$ does always hold. Though weaker than what the author of the article claims, this still sounds quite spectacular. I wonder if I'm not (still) overlooking something.
