How prove this inequality $(a+b)^rb^{2n-r}\le\dfrac{n}{4}+r,$ 
let $a,b>0$,and $n\ge 4$ be postive integers, such $(a+b)^{2n}=2n+\dfrac{n}{4},a^{2n}=\dfrac{n}{4}$
show that
  $$(a+b)^rb^{2n-r}\le\dfrac{n}{4}+r,\forall r\in[0,2n]$$

it seem hard for $n=4$ case.show this inequality $3^{\frac{n}{4}}\cdot (3^{\frac{1}{4}}-1)^{8-n}\le 1+n$
it is enought show that
$$((a+b)^{2n})^r(b^{2n})^{2n-r}\le\left(\dfrac{n}{4}+r\right)^{2n}$$
or 
$$(\dfrac{9n}{4})^{r}(b^{2n})^{2n-r}\le\left(\dfrac{n}{4}+r\right)^{2n}$$
it seem hard to prove it
 A: Following the hint given by guest, we consider $f(r):=(a+b)^rb^{2n-r}$, and we have to show that $f(r)\le \frac n4 + r$ for all $r\in[0,2n]$. 
We write $f(r)=\left(\frac ab +1\right)^rb^{2n}$, then $f'(r)=\left(\frac ab +1\right)^rb^{2n}ln\left(\frac ab +1\right)$ and 
$$
f''(r)=\left(\frac ab +1\right)^rb^{2n}\left(ln\left(\frac ab +1\right)\right)^2>0.
$$
So $f(r)$ is a convex funtion, hence, for all $t\in [0,1]$, we have
$$
f(t\cdot 2n+(1-t)\cdot 0)\le tf(2n)+(1-t)f(0).
$$ 
Now we have $f(2n)=\frac n4+2n$ and $f(0)=b^{2n}=\frac n4(3^{1/n}-1)^{2n}<\frac n4$. Setting $t:=\frac{r}{2n}$ we obtain
$$
f(r)=f(t\cdot 2n+(1-t)\cdot 0)\le \frac r{2n}\left(\frac n4+2n\right)+\left(1-\frac{r}{2n}\right)\frac n4=r+\frac n4,
$$
as desired.
A: Another version ...
From $a=\sqrt[2n]{\frac{n}{4}}$, $a+b=\sqrt[2n]{2n+\frac{n}{4}}$
we have $a\geq1$ and $a+b >1$, then:
$$(a+b)^{r}b^{2n-r}=\left(2n+\frac{n}{4}\right)^{\frac{r}{2n}}\left(b^{2n}\right)^{\frac{2n-r}{2n}} \tag{1}$$
However
$$b=\sqrt[2n]{2n+\frac{n}{4}}-\sqrt[2n]{\frac{n}{4}}<a \tag{2}$$
because 
$$\sqrt[2n]{2n+\frac{n}{4}}<2\sqrt[2n]{\frac{n}{4}} \Leftrightarrow \sqrt[2n]{\frac{9n}{4}}<2\sqrt[2n]{\frac{n}{4}} \Leftrightarrow \sqrt[2n]{9}<2, \forall n\geq2$$
Continuing $(1)$ and considering $(2)$:
$$(a+b)^{r}b^{2n-r}<\left(2n+\frac{n}{4}\right)^{\frac{r}{2n}}\left(a^{2n}\right)^{\frac{2n-r}{2n}}=\left(2n+\frac{n}{4}\right)^{\frac{r}{2n}}\left(\frac{n}{4}\right)^{\frac{2n-r}{2n}} \tag{3}$$
We will now look (considering $\ln{x}$ is concave function and $\frac{r}{2n}+\frac{2n-r}{2n}=1$) at:
$$\ln{\left(\left(2n+\frac{n}{4}\right)^{\frac{r}{2n}}\left(\frac{n}{4}\right)^{\frac{2n-r}{2n}}\right)}=\frac{r}{2n}\ln{\left(2n+\frac{n}{4}\right)}+\frac{2n-r}{2n}\ln{\left(\frac{n}{4}\right)}\leq \\
\ln{\left(\frac{r}{2n}\left(2n+\frac{n}{4}\right)+\frac{2n-r}{2n}\left(\frac{n}{4}\right)\right)}=\ln{\left(\frac{n}{4}+r\right)}$$
which, from $(3)$, leads to 
$$(a+b)^{r}b^{2n-r}<\left(2n+\frac{n}{4}\right)^{\frac{r}{2n}}\left(\frac{n}{4}\right)^{\frac{2n-r}{2n}} \leq \frac{n}{4}+r $$
