# determing constant in inequality with nonnegative numbers

Let $r \geq 1$ be an integer. Prove that there exists a constant $C_r = C(r)>0$ such that for any non-negative real numbers $a_1, a_2, \cdots, a_n \in [0, \infty)$ the following inequality holds:

$$(a_1 + a_2 + \cdots+ a_n)^r \leq C_r \left( \sum_{j=1}^n a_j^r + \sum_{1 \leq i_1 < i_2 <\cdots < i_r \leq n} a_{i_1} a_{i_2} \cdots a_{i_r} \right)$$.

Edit: I fixed a typo in the second summand. Now it is correct. Thank's ougao for pointing this out.

Edit2: Is it something trivial which I can't see? Because I can't understand why is negative voted.

I would really appreciate some ideas or anything that can help to prove this inequality.

• What do you think? – Did Jul 9 '14 at 17:53
• @Did: Unfortuetly I can't think anything... – passenger Jul 9 '14 at 17:56
• Something wrong in the second summand, there is only one term can appear in this notation – ougao Jul 9 '14 at 18:07
• @ougao: Yes you are right. The last number should be $a_{i_r}$. Thank's for the correction! I fix it right now. – passenger Jul 9 '14 at 18:10
• Look at examples where you have n=3 and r=1,2,3 what types of terms show up on the left and what terms show up on the right? – BeaumontTaz Jul 9 '14 at 23:44

I propose that the question has a typo and that the inequalities in the index of the summation cannot be strictly less than and need to be less than or equal to. Here is why.

Consider the case where $n=2$ and $r=3$. Clearly, the left hand side is an application of binomial theorem and we have $a_1^3+3a_1^2a_2+3a_1a_2^2+a_2^3$ The right hand side is (from the first term) at least $a_1^3 + a_2^3$. But the second summation has to use three $i$ values between $1$ and $2$ which it simply cannot do because $i\in\mathbb{N}$ and there are only $2$ such numbers (i.e. $1$ and $2$). Either you have to stipulate that $n\geq r$ or (and I think this is the way it should be) that the bounds are $1\leq i_1 \leq i_2 \leq \dots \leq i_r \leq n$. Also, we need to remove the cases where all $i$ are equal to avoid duplication of the first summation. (However, omitting the removal changes very little in the solution)

My reasoning for thinking that $n \geq r$ is not enough is shown in the example where $n=4$ and $r=3$. If you were to expand $(a_1+a_2+a_3+a_4)^3$ you would have terms of the form $a_i^3$, $a_i^2a_j$, and $a_ia_ja_k$ for $i\neq j\neq k$. However, on the right side, you'd only be able to get terms of the form $a_i^3$ from the first summation and terms of the form $a_ia_ja_k$ from the second summation. But we have terms of the form $a_i^2a_j$ on the left that aren't on the right. Which means we can't always pick a constant that is bigger than $a_i^2a_j$ for any $a_i,a_j$.

Therefore, I will change the second summation to this

$$\sum_{\begin{array}{c} 1\leq i_1 \leq i_2 \leq ... \leq i_r \leq n \\ \neg (i_1=i_2=...=i_r)\end{array}}a_{i_1}a_{i_2}\cdots a_{i_r}$$

So we want to show that for $r\in\mathbb{Z}$ such that $r \geq 1$, $\exists\;C_r$ such that for any $a_1, a_2, \ldots, a_n \in \left[0,\infty\right)$

$$\left(a_1 + a_2 + \ldots + a_n\right)^r \leq C_r\left(\sum_{i=1}^na_i^r+\sum_{\begin{array}{c} 1\leq i_1 \leq i_2 \leq ... \leq i_r \leq n \\ \neg (i_1=i_2=...=i_r)\end{array}}a_{i_1}a_{i_2}\cdots a_{i_r}\right)$$

Consider

\begin{align} \left(a_1 + a_2 + \ldots + a_n\right)^r &= \underbrace{\left(a_1 + a_2 + \ldots + a_n\right)}_1 \underbrace{\left(a_1 + a_2 + \ldots + a_n\right)}_2 \cdots \underbrace{\left(a_1 + a_2 + \ldots + a_n\right)}_r \\ & =\sum_{1\leq i_1,i_2,\ldots,i_r \leq n}a_{i_1}a_{i_2}\cdots a_{i_r}\end{align}

Which is all the products of grabbing one element from the first parenthesis, one element from the second, and so on through the $r$th parenthesis.

We can now group like terms and order the $i$'s. We can do this because of the commutivity of multiplication.

$$\left(a_1+a_2+\ldots + a_n\right)^r = \sum_{1\leq i_1 \leq i_2 \leq \cdots \leq i_r \leq n}c_ka_{i_1}a_{i_2}\cdots a_{i_r}$$

Where $c_k$ is the number of times that particular term shows up. Let $C_r$ be the largest $c_k$. Therefore, $c_k \leq C_r \;\forall \;k$.

$$\left(a_1+a_2+\ldots + a_n\right)^r \leq C_r\sum_{1\leq i_1 \leq i_2 \leq \cdots \leq i_r \leq n}a_{i_1}a_{i_2}\cdots a_{i_r}$$

Now it's clear that by (unnecessarily) breaking up the summation into two cases, one where all $i$'s are equal and one where all $i$'s cannot be equal we have

$$\left(a_1+a_2+\ldots + a_n\right)^r \leq C_r\left(\sum_{i=1}^na_i^r+\sum_{\begin{array}{c} 1\leq i_1 \leq i_2 \leq ... \leq i_r \leq n \\ \neg (i_1=i_2=...=i_r)\end{array}}a_{i_1}a_{i_2}\cdots a_{i_r}\right)$$

We know that $C_r > 0$ because for the non-trivial case where at least one $a_i>0$, both the left hand side and the parenthesis on the right hand side are positive.

• Thank you vey much for your reply! – passenger Jul 10 '14 at 7:44