(1) $-2\leq a_{i} \leq 2$ $~(i=1,2,3,4,5)$
(2) $\displaystyle\sum_{cyclic}a_{i}=0$
then, find the maximum value of $\displaystyle\sum_{cyclic}a_{i}^{3}$
also, can it be generalized as for $\displaystyle\sum_{cyclic}a_{i}^{n}$ $~(n\in N)$
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$\begingroup$ @Tim Ratigan/ 2,2, $-\dfrac{4}{3}$, $-\dfrac{4}{3}$, $-\dfrac{4}{3}$ : it is greater than $6$. $\endgroup$– chloe_shiDec 25, 2013 at 10:27
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$\begingroup$ What do you mean by summing over cyclic? $\endgroup$– NewbDec 25, 2013 at 10:35
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$\begingroup$ Woops. I thought they had to be integers for some reason @chloe_shi $\endgroup$– Tim RatiganDec 25, 2013 at 10:56
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$\begingroup$ @Newb summing over cyclic means $a_1+a_2+a_3+a_4+a_5$ in this case. $\endgroup$– lspDec 25, 2013 at 11:52
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$\begingroup$ Lagrange multiplier? $\endgroup$– DeepSeaDec 25, 2013 at 12:05
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1 Answer
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The function $f(u) = u^3$ is convex for $u \ge 0$ and concave for $u \le 0$. From this we have for $u, v \in \{ a_i \}$:
$$ f(u) + f(v) \le \begin{cases} 2f\left(\dfrac{u+v}2\right) & \qquad u, v \in [-2,0] \\ f(2) + f(u+v-2) & \qquad u, v \in [0, 2] \end{cases} \tag{1}$$
Also, we have, if $ u < 0 < v$: $$ f(u) + f(v) \le \begin{cases} f(0) + f(u+v) & \qquad u+v \le 0 \\ f(2) + f(u+v-2) & \qquad u+v \ge 0 \end{cases} \tag{2}$$
WLOG, we can assume $a_i$ are ordered in descending order. Let the first $k$ terms be positive. Then using the reasoning in $(1)$ above we have $$\sum_{i=1}^k f(a_i) \le (k-1)f(2) + f\left(\sum_{i=1}^k a_i - 2(k-1) \right)$$ Thus it is clear that we can set $k-1$ of these variables to $2$ for maximising the above partial sum. Let the remaining positive variable be $a_k$.
Similarly, for the terms which are $\le 0$, we have from $(1)$: $$\sum_{i=k+1}^5 f(a_i) \le (5-k)f\left(\frac1{5-k} \sum_{i=k+1}^5 a_i \right) $$ hence we can replace all such terms with their arithmetic mean $\mu$ to maximise this partial sum.
Now we have $\mu < 0 < a_k$ and hence using $(2)$ we can replace $(\mu, a_k)$ with $(2, \mu+a_k-2)$ or $(0, \mu+a_k)$ to increase the sum (depending on the sign of $\mu + a_k$), thereby making all positive terms equal to $2$. So we have that the maximum is when some $k$ of the $a_i$s are $2$, and the remaining are all equal. Obviously $\mu = -\dfrac{2k}{5-k}$, giving a sum of $$g(k) = kf(2) - (5-k)f\left(\frac{2k}{5-k}\right) = 8k - \frac{(2k)^3}{(5-k)^2} $$ It is easy to check that $g'(k) = 0$ when $k = \frac53$, and that $k = 2$ gives the maximum among integers in $[1, 5]$.
The method can be adapted in general for real $a_i \in [c, d]$, and natural $i \le m$ to find the maximum of $\sum a_i^n$ subject to $n \in \mathbb{N}$ and $\sum a_i = 0$.
