find the maximum and minimum of $\sum_{i=1}^{n} (10x^3_{i}-9x^5_{i})$ 
Let $x_{i}\ge 0$ such that $$x_{1}+x_{2}+\cdots+x_{n}=1.$$
Find the maximum and minimum of
$$f=10\sum_{i=1}^{n}x^3_{i}-9\sum_{i=1}^{n}x^5_{i}.$$

I have proved $n=2$
$$1\le f\le\dfrac{9}{4}$$ see: wolfarma
When $n=3$,  How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$
So I suspect that the general positive integer $n\ge 4$, also has the following conclusion
$$1\le f\le \dfrac{9}{4}$$
when $$(x_{1},x_{2},\cdots,x_{n})=(1,0,0,\cdots,0),f=1$$
and
$$(x_{1},x_{2},\cdots,x_{n})=(0,\dfrac{1}{2}+\dfrac{1}{2\sqrt{3}},\dfrac{1}{2}-\dfrac{1}{2\sqrt{3}},0,\cdots,0),f=\dfrac{9}{4}$$But how to prove or reverse this conclusion?
 A: Assumption
If $f(x)$ is convex strictly increasing for $0 \le x \le 1$ then
$$
\min\sum_{k=1}^n f(x_k) \ \ \mbox{s. t. }\ \ \sum_{k=1}^n x_k = 1, \ \ x_k > 0
$$
has it's minimum at $ x_1 = \cdots = x_n = \frac 1n$
Using Lagrange Multipliers the problem can be stated as
$$
L(x,\lambda) = \sum_{k=1}^n f(x_k)-\lambda\left(\sum_{k=1}^n x_k -1\right)
$$
so the stationary points are the solutions for
$$
\frac{d}{dx_k}f(x_k) -\lambda = 0\\
\sum_{k=1}^n x_k - 1 = 0
$$
or 
$$
x_k = \frac{\sqrt{5\pm\sqrt{5} \sqrt{5-\lambda }}}{\sqrt{15}}
$$
now assuming all $x_k =  \frac{\sqrt{5+\sqrt{5} \sqrt{5-\lambda }}}{\sqrt{15}}$ 
so
$$
\lambda = \frac{15(2n^2-3)}{n^4}
$$
then
$$
x_k =\left\{\frac 1n,\frac{\sqrt{\sqrt{\frac{\left(n^2-3\right)^2}{n^4}}+1}}{\sqrt{3}}\right\}
$$
The last value is discarded because does not observe the restriction then we follow with $x_k = \frac 1n$ so
$$
\min \sum_{k=1}^n f(x_k) = n f\left(\frac 1n\right)
$$
Attached a plot showing the $F_n(x_n^*) = \sum_{k=1}^n f(x_n^*)$ evolution assuming $n$ continuous
NOTE
$$
f(x) = 10x^3-9 x^5
$$
is convex strictly increasing for $0\le x \le 0.5 $ so for $n \gt 2$ we have at $x^* = \frac 1n$ a local minimum.

For $n = 2$ making $x_2=\lambda x_1, \ x_3 = \mu x_1$ we have
$$
\min_{\lambda,\mu}\frac{10(1+\lambda^3+\mu^3)}{(1+\lambda+\mu)^3}-\frac{9(1+\lambda^5+\mu^5)}{(1+\lambda+\mu)^5}
$$
which gives the feasible solution
$$
x_1 = x_2 = \frac 12
$$
A: For $n\geq4$ the minimal value is not $1$.
For example, take $n=4$ and $x_1=x_2=x_3=x_4=\frac{1}{4}.$
We'll prove that for $n\geq3$ the minimal value it's $\frac{10n^2-9}{n^4}$ and occurs for $x_1=x_2=...=x_n=\frac{1}{n}.$
Indeed, we need to prove that
$$\sum_{k=1}^ng(x_k)\geq ng\left(\frac{\sum\limits_{k=1}^n}{n}\right),$$
where $g(x)=10x^3-9x^5,$ which has unique inflection point $x_0=\frac{1}{\sqrt3}$ on $[0,1]$.
Thus, by Vasc's HCF Theorem it's enough to prove our inequality for
$x_1=x_2=...=x_{n-1}=a$ and $x_n=1-(n-1)a,$ which gives
$$(na-1)^2(9n^3(n^3-4n^2+6n-4)a^3-9n^2(3n^3-7n^2+3n+3)a^2+n(17n^3+2n^2-18n-18)a+n^3+n^2-9n-9)\geq0,$$ which is true because
$$9n^3(n^3-4n^2+6n-4)a^3-9n^2(3n^3-7n^2+3n+3)a^2+$$
$$+n(17n^3+2n^2-18n-18)a+n^3+n^2-9n-9\geq0$$ for all $n\geq3$ and $0\leq a\leq\frac{1}{n-1}.$
The proof of the last statement for you.
The hint for the maximal value.
Since $g$ has an unique inflection point on $[0,1]$, 
we can use Karamata and Jensen and we can get an inequality of one variable.
For $n=4$ it works very well, but for $n\geq5$ it's very ugly.  
