For non-negative real numbers $x_{1},\: x_{2},\: ...,\: x_{n}$ , which satisfy $x_{1}+x_{2}+...+x_{n}=1$ , find the largest possible value of $\displaystyle \sum_{j=1}^{n}(x_{j}^{4}-x_{j}^{5})$ .

Give your answer to 3 significant figures.

The answer is 0.0833.

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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

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