I have tried doing this exercise,
Let $m,n\in\mathbb{N}, m\leq n$, prove that
$$\log\left(\frac{4^n}{\displaystyle\sqrt{2n+1}{2n\choose n+m}}\right)\geq \frac{m^2}{n}$$
I achieved some results like for example,
$$\displaystyle\sum_{i=0}^n 2^i\binom{2n-i}{n} = 4^n$$ and
$$\displaystyle {{2n}\choose{n}} > \frac{4^n}{2n}$$ trying to find a relationship but it doesn't work for me. Any idea?
Best Answer
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: $$\log\left(\frac{4^n}{\sqrt{2n+1}}\right)\ge\frac{m^2}n+\log\binom{2n}{n+m}=f_n(m).$$ 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 $z\ne-1,-2,\dots$, there is an equation for the digamma function, namely $$\psi(z+1)=-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+z}\right),$$ 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 $$-g_n(m)=\psi(n-m+1)-\psi(n+m+1)=\sum_{k=1}^\infty\left(\frac1{k+n+m}-\frac1{k+n-m}\right).$$
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 $$\frac d{dm}g_n(m)=\sum_{k=1}^\infty\left(\frac1{(k+n+m)^2}-\frac1{(k+n-m)^2}\right)>0$$ for all $m$. This means, in particular, that if $m$ is not an integer, then $g_n(m)$ is sandwiched between $g_n(\lfloor m\rfloor)$ and $g_n(\lceil m\rceil)$. Obviously, the function $\frac{2m}n$ is increasing with respect to $m$. This all implies that it is sufficient to show that $$\tag{*}g_n(m)\ge\frac{2m-2}n$$ for integers $m\ge\frac n2$.
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 $m=\lceil\frac n2\rceil$. But then \begin{align*}g_n(m)&\ge\sum_{k=1}^n\frac1{k+(n-1)/2}\\\frac{2(m-1)}n&\le1.\end{align*} So it suffices to prove that $h(n)=\sum_{k=1}^n\frac1{k+(n-1)/2}\ge1$.
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 $n\to\infty$, 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 $h(n)\to\log3$ 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)