<s>[INST] 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?
 [/INST] 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.  
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