For what percentage of numbers does this proof of Goldbach's conjecture hold? Question
For what percentage of numbers does the below inequality hold?
$$ \pi(2m) > \frac{\phi(2 m) -1}{2} $$
where $m$ is not a prime or $1$, $\pi(m)$ is the number of primes less than $m$ and $\phi$ is the Euler totient function.
Background + Proof
I was trying to prove Goldbach's Conjecture are realised I could do it for certain numbers.
Consider the following where $m$ is not a prime:
$$ (2m)! =  2m^2 (m^2 -1)(m^2 -4)(m^2 -9) \dots (m^2 -(m-1)^2)$$
Notice, $(2m)!$ will contain primes of the form $p_k =m \pm n \leq 2m$.  The number of such primes are $\pi(2m)$. Further, the $m$ and $n$ must be co-prime. Then the number of possible "slots" of the form $m \pm n$ which can house primes (disregarding $p_1 = 2$) are:
$$ S(m) =  \phi(2 m) -1 $$
To derive the above the following are considered:
$1$. If $m$ is even then $(m^2 - (\text{even})^2) = \text{even}$
$2$. $m$ is not a prime as mentioned before.
$3$. $m - (m-1)$ is cannot be a prime.
Now, if:
$$ \pi(2m) > \frac{S(m)}{2} $$
Then we can prove Goldbach's conjecture for that number as that would imply one of the slots $m+n = p_i$ has a corresponding slot housing a prime $m+n =p_k$. And thus, if add both of them:
$$ 2m = p_i + p_k$$
Example
Consider $m = 15 = 3 \cdot 5$. After considering $30!$ eliminating "slots":
$$ S(15) =  (1-\frac{1}{3})(1-\frac{1}{5}) 30 - 1= 15 $$
where the remaining slots are:
$$ (m^2 -1),(m^2 -7^2),(m^2 -11^2),(m^2 -13^2),(m^2 -17^2),(m^2 -19^2),(m^2 -23^2),(m + 29) $$
However, the number of primes are $\pi(30) = 10$. Since, there are more primes than "slots" available for them. Then one of the slots of the form $m$.
 A: (Note, I didn't read the background, just the question.)
One has 
$$\pi(2x) \sim \frac{2x}{\log x}$$
by the prime number theorem and
and
$$\phi(x) \gg \frac{e^{-\gamma} x}{\log \log x}$$
by standard estimates (see the wikipedia page https://en.wikipedia.org/wiki/Euler%27s_totient_function), so since the latter grows much faster there can only be finitely many integers satisfying your inequality (and this finite set has density zero, obviously). More precisely, one has:
$$\frac{n}{\phi(n)} \le e^{\gamma} \left( \log \log n + \frac{2.5}{e^{\gamma} \log \log n} \right)$$
for all $n \ge 3$ except for $n = 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23$,
(See Lemma 4 here: https://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/grytczuk.pdf)
and
$$\pi(x) < 1.25506 \cdot \frac{x}{\log x}$$ 
for $x \ge 17$ (see https://en.wikipedia.org/wiki/Prime-counting_function) and just from these inequalities (computing $\phi(223092870)$ by hand) we already get
$$\frac{\phi(2x) - 1}{2} > \pi(2x), \quad x > 10^6$$
But then the smaller examples one can check by computer, and find the inequality you want holds for exactly $649$ integers, the largest one being $45045$ with
$$\frac{\phi(90090) - 1}{2} = 8639.5 < 8726 = \pi(90090).$$
A: We can rephrase your question as: for which $m$ does
$$
\pi(2m) \geq \frac{\phi(2 m)}{2}
$$
hold, or equivalentely
$$
\frac{\phi(2m)}{\pi(2m)} \leq 2.
$$
Now for large enough $m$, $\pi(m) \approx \frac{m}{\log(m)}$, so this inequality becomes (roughly speaking)
$$
\frac{\phi(2m)\log(m)}{m}  \leq 2.
$$
One expression for $\phi(k)$ is $k\prod_{p \mid k}\left(1-\frac1p\right)$, where the variable $p$ runs over the primes, so we can rewrite this as
$$
\log(m)\prod_{p \mid 2m}\left(1-\frac1p\right) \leq 1.
$$
Taking logarithms on both sides,
$$
\log(\log(m)) + \sum_{p \mid 2m}\log\left(1-\frac1p\right) \leq 0.
$$
For large $p$, $-\frac1p$ is a good approximation for $\log\left(1-\frac1p\right)$, so the question becomes
$$
\log(\log(m)) - \sum_{p \mid 2m} \frac1p \leq 0.
$$
Now a standard approximation tells us that this last sum is approximately equal to $\log(\log(p))$ where $p$ is the largest prime in the sum. Clearly the largest prime in this sum will be much smaller than $m$, because we took the best case $m$: where $m$ is a product of lots of distinct small primes.
Thus your inequality should hold almost never.
