A 1999 IMO Chinese Team Selection Question

We can smooth the inequality (thanks to Challenge Master, I originally called it "the adjustment method", which was barely professional), i.e. proving when there are fewer non-zero numbers, the value of sum can get larger. To do this, first compare the value of $(x+y)^{4}-(x+y)^{5}$ and $x^{4}-x^{5}+y^{4}-y^{5}$ :

$(x+y)^{4}-(x+y)^{5}-(x^{4}-x^{5}+y^{4}-y^{5})$

(note: when this is larger than $0$ , then we know that converting two numbers into one can make the result larger)

$=xy(4x^{2}+6xy+4y^{2})-xy(5x^{3}+10x^{2}y+10xy^{2}+5y^{3})$

$=\frac{1}{2}xy(7x^{2}+12xy+7y^{2}+x^{2}+y^{2})-5xy(x^{3}+2xy^{2}+2x^{2}y+y^{3})$

$\geq \frac{7}{2}xy(x^{2}+2xy+y^{2})-5xy(x^{3}+3xy^{2}+3x^{2}y+y^{3})$

$=\frac{1}{2}xy(x+y)^{2}[7-10(x+y)]$ ,

only if $x,y> 0,\; x+y< \frac{7}{10}$ , the above equation is larger than 0.

If the non-zero numbers in $x_{1},x_{2},\: ...,x_{n}$ is less than two, then the value of the required sum is $0$ .

The following observes when there are two or more non-zero numbers in $x_{1},x_{2},\: ...,x_{n}$ .

If some three numbers $x_{i},x_{j},x_{k}> 0$ , then it is certain that the sum of two of these numbers is less or equal to $\frac{2}{3}< \frac{7}{10}$ . We can convert these two numbers into one, and as the above has demonstrated, the required sum gets larger. After some limited adjustments, eventually we have only 2 non-zero integers left, which we can assume them to be $x,y> 0,\; x+y=1$ .

Then $x^{4}-x^{5}+y^{4}-y^{5}=x^{4}(1-x)+y^{4}(1-y)=x^{4}y+y^{4}x$

$=xy(x^{3}+y^{3})=xy[(x+y)^{3}-3xy(x+y)]=xy(1-3xy)$ ,

of which the value is the largest when $xy=\frac{1}{6}$ , which is

$\frac{1}{6}-3\times (\frac{1}{6})^{2}=\boxed{\frac{1}{12}}$ .

Let $f(x)=x^4-x^5\\$ $f'(x)=4x^3-5x^4$ where $f'(x)$ denotes the first derivative of $f(x)$ $f''(x)=12x^2-20x^3=4x^2(3-5x)$

$f''(x)>0$ when $(-\infty,\frac{3}{5})$ and hence convex in this interval

$f''(x)<0$ when $(\frac{3}{5},\infty)$ and hence concave in this interval

Suppose $x_1,x_2,...,x_i$ are in the interval in which $f(x)$ is convex and $x_{i+1},...x_n$ are in the interval in which it is concave.

$f(x)$ is maximum when $n-i$ is maximum. Since $\displaystyle \sum_{i=1}^n =1$ and interval in which $f(x)$ is convex is $(\frac{3}{5},\infty)$ So maximum value that $n-i$ attains is $1$

Maximum of $f(x)$ occurs when $x_n=x_{n-1}=...=x_{i+1}, x_1=1-(n-i)x_n$ and $x_2=x_3=...=x_i=0$

This above step is intutively evident by the properties of convexity and concavity.

The required sum in the question becomes $S=(n-i)(x^4_n-x^5_n)+(1-(n-i)x_n)^4-(1-(n-i)x_n)^5$ Put $n-i=1$ as concluded above and we need to maximize the function $S(x)=x^4-x^5+(1-x)^4+(1-x)^5$ $=x^4-x^5+(1-x)^4(1-(1-x))$ $=x^4-x^5+(1-x)^4(x))$ $=x(x^4-x^5+(1-4x^3+6x^2-4x+x^4)$ $=-3x^4+6x^3-4x^2+x$ $S'(x)=-12x^3+18x^2-8x+1=(1-2x)(6x^2-6x+1)$

So max occurs when $x_1=\frac{3+\sqrt{3}}{6}, x_2=\frac{3-\sqrt{3}}{6}$ and all others are $0$ . These $x_1$ and $x_2$ are actually the roots of $6x^2-6x+1$ . Finally putting in $S$ the required maximum value is $\boxed{\frac{1}{12}}$