Proving $logleft(frac{4^n}{sqrt{2n+1}{2nchoose n+m}}right)geq frac{m^2}{n}$

With @skbmoore's work, we know that this is true for $m<sqrt{log(pi/2)}n$. I'll now show that it's also true for $m>frac12n$, which will obviously prove the result.

Rearrange the desired inequality as thus: $$logleft(frac{4^n}{sqrt{2n+1}}right)gefrac{m^2}n+logbinom{2n}{n+m}=f_n(m).$$ We're basically going to try to show that $f_n(m)$ is decreasing after $m=n/2$ (actually, it's a bit before that; I believe it is somehow related to OEIS A143978).

Observe that $$frac d{dm}f_n(m)=frac{2m}n+psi(n-m+1)-psi(n+m+1),$$ where $psi$ is the digamma function. (This is from Wolfram Alpha; I've actually never worked with $psi$ before today, so please let me know if I mess up anywhere here—I'm a bit out of my depth!) Do note that we're extending $f_n(m)$ to be over $[1,n]$, instead of only the integers.

Apparently, for $zne-1,-2,dots$, there is an equation for the digamma function, namely $$psi(z+1)=-gamma+sum_{k=1}^inftyleft(frac1k-frac1{k+z}right),$$ where $gamma$ is the Euler-Mascheroni constant. It is fortunate for us, then, that $n-m$ and $n+m$ are never nonnegative integers! This means, in particular, that $$-g_n(m)=psi(n-m+1)-psi(n+m+1)=sum_{k=1}^inftyleft(frac1{k+n+m}-frac1{k+n-m}right).$$

Our goal, then, is going to be to show that $g_n(m)>frac{2m}n$ for all $n>m>frac n2$. Then we can show that $f_n(m)=frac{2m}n-g_n(m)<0$.

First, observe that $$frac d{dm}g_n(m)=sum_{k=1}^inftyleft(frac1{(k+n+m)^2}-frac1{(k+n-m)^2}right)>0$$ for all $m$. This means, in particular, that if $m$ is not an integer, then $g_n(m)$ is sandwiched between $g_n(lfloor mrfloor)$ and $g_n(lceil mrceil)$. Obviously, the function $frac{2m}n$ is increasing with respect to $m$. This all implies that it is sufficient to show that $$tag{*}g_n(m)gefrac{2m-2}n$$ for integers $mgefrac n2$.

However, for integers $m$, we know that $g_n(m)$ telescopes as $$g_n(m)=sum_{k=1}^{2m}frac1{k+n-m}.$$ Now, if $(*)$ holds for $m$, then it holds for $m+1$. This can be seen by observing that the left hand side increases by $frac1{n-m}+frac1{n+m+1}$, while the right side increases by $frac2n$.

Thus it suffices to prove the statement $(*)$ for $m=lceilfrac n2rceil$. But then begin{align*}g_n(m)&gesum_{k=1}^nfrac1{k+(n-1)/2}\frac{2(m-1)}n&le1.end{align*} So it suffices to prove that $h(n)=sum_{k=1}^nfrac1{k+(n-1)/2}ge1$.

But it is easy to see that if we define $h(n)$ to be the sum above, but removing the floors, then $frac d{dn}h(n)<0$. Moreover, as $ntoinfty$, this approaches $log3>1$, according to Wolfram. If someone wants to give me a tip as to how to actually show this limit, I'd love to hear it, but, honestly, I'm a bit pooped! (Check out @skbmoore's explanation for why $h(n)tolog3$ in the comments!)

This does, however, prove the conjecture! I'm sure there's a much simpler way to do this, since there's no real intuition here; it's just bashing with the one tool I know how to use (Wolfram Alpha! :D)

Here is a partial proof. I have some ideas, but it will be a while before I can return to it. I'll show that the conjecture is true for $m<sqrt{log{(pi/2)}} n sim .672 n.$ Maybe someone else can use these ideas for a full proof.

Use the fact that the central binomial $binom{2n}{n+m}$ has its max at $m=0.$ Do an asymptotic expansion

$$ binom{2n}{n+m} big/binom{2n}{n}=1-frac{m^2}{n}+frac{m^2(1+m^2)}{2n^2}-frac{m^2(1+4m^2+m^4)}{6n^3}... $$ It 'just so happens' that these are the first three terms in $$exp{big(-frac{m^2}{n}(1-frac{1}{2n}) big)} =1-frac{m^2}{n}+frac{m^2(1+m^2)}{2n^2}-frac{m^2(3m^2+m^4)}{6n^3}...$$ match. (The gaussian approximation is well known and I added the factor $(1-1/(2n))$ to match the third term.) The exponential makes a convenient bound to go in this problem (note the flip): $$ binom{2n}{n} big/binom{2n}{n+m} ge exp{big(frac{m^2}{n}(1-frac{1}{2n}) big)} $$

Then $$L:=logBig(frac{4^n}{sqrt{2n+1}binom{2n}{n+m}}Big)= logBig(frac{4^n}{sqrt{2n+1}binom{2n}{n}} binom{2n}{n} big/binom{2n}{n+m}Big)$$ $$ gelogBig(frac{4^n}{sqrt{2n+1}binom{2n}{n}} exp{big(frac{m^2}{n}(1-frac{1}{2n})}big) Big) $$ $$ geq frac{1}{2} log{big( pi n/(2n+1) big)} + frac{m^2}{n}(1-frac{1}{2n}) $$ where the Stirling approximation has been used for the central binomial. For large $n$ proposer's reduces to $$ frac{1}{2} log{big( pi /2)} > frac{m^2}{2n^2} .$$ This is indeed true for $m<sqrt{log{(pi/2)}} n.$

The problem that I see with this method is that the Gaussian approximation, even with my correction to get the third order term matched, does not do a good job in the 'wings' (large $m.$) A better function is needed, and I believe there are 'entropy' function formulations that can do this. I don't know if an analytic solution will be available using it, but at least the one I've given gets part of the way there.