(Solution of the modest subproblem)

Taking the second derivative we see that it equals $(1+x+\dots+x^{k-1})^{-2}f(x)$, where $f_k(x)$ is a polynomial (not a surprise) with explicit formulae for coefficients: $$ f_k(x)=\sum_{n=0}^{k-3}\binom{n+3}3x^n+\sum_{n=k-2}^{2k-4}\left(\binom{2k-n-1}3-k(2k-n-3)\right)x^n. $$ It is straightforward that the sum of coefficients of $f_k$ equals $k^2(k-1)(k-5)/12\geqslant 0$ when $k\geqslant 5$, and negative coefficients go with larger powers of $x$, thus $f_k$ is non-negative on $[0,1]$.

maybe such brute force approach works for other questions too.

(Solution of the modest subproblem)

The second derivative of $g_k(x)$ equals $(1+x+\dots+x^{k-1})^{-2}f_k(x)$, where $f_k(x)$ is a polynomial (not a surprise) with explicit formulae for coefficients: $$ f_k(x)=\sum_{n=0}^{k-3}\binom{n+3}3x^n+\sum_{n=k-2}^{2k-4}\left(\binom{2k-n-1}3-k(2k-n-3)\right)x^n. $$ It is straightforward that the sum of coefficients of $f_k$ equals $k^2(k-1)(k-5)/12\geqslant 0$ when $k\geqslant 5$, and negative coefficients go with larger powers of $x$, thus $f_k$ is non-negative on $[0,1]$.

maybe such brute force approach works for other questions too.

(Solution of the modest subproblem)

Taking second derivative we see that it equals $(1+x+\dots+x^{k-1})^{-2}f(x)$, where $f_k(x)$ is a polynomial (not a surprise) with explicit formulae for coefficients: $$ f_k(x)=\sum_{n=0}^{k-3}\binom{n+3}3x^n+\sum_{n=k-2}^{2k-4}\left(\binom{2k-n-1}3-k(2k-n-3)\right)x^n. $$ It is straightforward that the sum of coefficients of $f_k$ equals $k^2(k-10(k-5)/12\geqslant 0$ when $k\geqslant 5$, and negative coefficients go with larger powers of $x$, thus $f_k$ is non-negative on $[0,1]$.

maybe such brute force approach works for other questions too.

(Solution of the modest subproblem)

Taking the second derivative we see that it equals $(1+x+\dots+x^{k-1})^{-2}f(x)$, where $f_k(x)$ is a polynomial (not a surprise) with explicit formulae for coefficients: $$ f_k(x)=\sum_{n=0}^{k-3}\binom{n+3}3x^n+\sum_{n=k-2}^{2k-4}\left(\binom{2k-n-1}3-k(2k-n-3)\right)x^n. $$ It is straightforward that the sum of coefficients of $f_k$ equals $k^2(k-1)(k-5)/12\geqslant 0$ when $k\geqslant 5$, and negative coefficients go with larger powers of $x$, thus $f_k$ is non-negative on $[0,1]$.

maybe such brute force approach works for other questions too.