Consider only $k>4$$k>2$. Denote $$f(x)=x\cdot \frac{(1-x)^{k-1}}{(k+1)^{k-2}}+\frac{(1-2x)^k}{k^k},$$ we need to prove that $f(x)\leqslant f(\frac1{k+2})$ for $x\in [0,1]$. We have $f'(\frac1{k+2})=0$ and $$f'(0)=(k+1)^{2-k}-2k^{1-k}=k^{2-k}\left(\left(1+\frac1k\right)^{2-k}-\frac2k\right)>0$$ by Bernoulli inequality $(1+x)^a>1+ax$ for $a=2-k$, $x=1/k$. Also $f(1)$ equals $\pm f(0)$. It means that the maximal value of $f$ on $[0,1]$ is attained at an interior point $a\in (0,1)$, and thus $f'(a)=0$. A technical fact is that $a$ can not belong to $[1/2,1]$ (in general, for large $k$ this is simply because LHS is too small).
Having this in mind, let me show how to finish the proof. I claim that the equation $f'(x)=0$$f'$ has unique solutionroot (multiplicity counted) on $[0,1/2]$$(0,1/2]$, and this root is $\frac1{k+2}$, soand at most one root on $a=\frac1{k+2}$$[1/2,1)$. In both cases the only possible maximum point is $\frac{1}{k+2}$ (the second extremal point would be a local minimum)
We have $$f'(x)=(1-x)^{k-2}(1-kx)(k+1)^{2-k}-2(1-2x)^{k-1}k^{1-k}=0.$$ Denote $y=\frac{1-x}{1-2x}\in (1,\infty)$$y=\frac{1-x}{1-2x}$, then $x=\frac{1-y}{1-2y}$, $1-kx=\frac{(k-2)y-(k-1)}{1-2y}$. Our equation $f'(x)=0$ in terms of $y$ rewrites as $$y^{k-2}((k-2)y-(k-1))(k+1)^{2-k}+2k^{1-k}=0.$$ LHS is monotemonotone in $y$ for $y\in (1,\infty)$ (corresponds to $x\in (0,1/2)$) and for $y<0$ (corr. $x\in (1/2,1)$), so there is indeed at most one solutionthat implies the above claim.