# How to show a scalar inequality?

Let $a_1,\cdots, a_n\ge0$ and $S=a_1+\cdots+a_n$ with $k\ge 2, k\in\mathbb{Z}$. How to show $$\sum_{i=1}^n(S-a_i)^k\le 2^{k-2}\left(\sum\limits_{i=1}^n a_i ^k+(n-2)S^k\right)?$$

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There must be a slick way to do this, but here's one way. We will split the problem into different cases; the inequality is quite obvious in each of the cases. (Convince yourself that the cases exhaust all the possibilities. :))

Case 1: $n = 2$. For $n=2$, we have $S = a_1 + a_2$. In this case the inequality is obvious: $$(S - a_1)^k + (S - a_2)^k = a_2^k + a_1^k,$$ which is clearly at most the right hand side. So we will assume $n \geq 3$ from now on.

Case 2: $k = 2$, $n$ arbitrary. For $k=2$, there's equality: $$\sum_{i} (S - a_i)^2 = nS^2 - 2S \sum_i a_i + \sum_i a_i^2 = nS^2 - 2S^2 + \sum_i a_i^2= (n-2) S + \sum_i a_i^2,$$ which is also the right hand side.

Case 3: $k=3, n=3$. In this case, the left hand side is: $$\begin{eqnarray*} && (a_2+a_3)^3 + (a_3+a_1)^3 + (a_1+a_2)^3 \\ &=& 2(a_1^3+a_2^3+a_3^3) + 3(a_1^2a_2+a_1a_2^2+ a_2^2 a_3+a_2a_3^2+ a_3^2a_1+a_3a_1^2), \end{eqnarray*}$$ whereas the right hand side is $$2(a_1^3+a_2^3+a_3^3) + 2(a_1+a_2+a_3)^3 .$$ Expanding the right hand side by the multinomial theorem, it is easy to check that the right hand side dominates the left term-by-term, and hence it is greater than or equal to the left.

Case 4: $k = 3, n \geq 4$. The only remaining case is when $k = 3$ and $n \geq 3$. Collect the terms to the right: $$\begin{eqnarray*} && 2(\sum_i a_i^3 + (n-2)S^3) - \sum_i (S - a_i)^3 \\ &=& (2n-4)S^3 + 2 \sum_i a_i^3 - n S^3 + 3S^2 \sum_i a_i - 3 S \sum_i a_i^2 + \sum_i a_i^3 \\ &=& (n-4)S^3 + 3 \sum_i a_i^3 + 3S^2 \cdot S - 3 S \sum_i a_i^2 \\ &\geq& 3S^3 + 3 \sum_i a_i^3 - 3 S \sum_i a_i^2 \\ &\geq& 3S^3 - 3 S \sum_i a_i^2 \\ &\geq& 3S^3 - 3 S \left(\sum_i a_i \right)^2 \\ &\geq& 3S^3 - 3 S \cdot S^2 = 0. \end{eqnarray*}$$

Case 5: $k \geq 4, n \geq 3$. Similarly, for $k \geq 4$ (and $n \geq 3$), the inequality is again obvious: $$\sum_i (S - a_i)^k \leq n S^k \leq 4(n-2) S^k \leq 2^{k-1}(n-2)S^k,$$ which is clearly smaller than the right hand side. This leaves us with a narrow range of parameters to check the inequality.

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I think it is possible to do this without case analysis (or may be with just 1-2 cases), but nothing comes to mind as yet. – Srivatsan Sep 23 '11 at 3:56