How to prove that $\left(\sum\limits_{k=1}^{n}a_{k}\right)^2\ge\sum\limits_{k=1}^{n}a_k^3$?

Question

Let $$a_{n}\ge a_{n-1}\ge\cdots\ge a_{0}= 0,$$ and for any $i,j\in\{0,1,2\dots,n\},j>i$, there is $$a_{j}-a_{i}\le j-i.$$ Prove that $$\left(\sum_{k=1}^n a_k \right)^2\ge\sum_{k=1}^n a_k^3.$$

My idea is by mathematical induction:

Assume that $n$ is true, meaning $$\left(\sum_{k=1}^n a_k\right)^2\ge\sum_{k=1}^n a_k^3,$$

then for $n+1$, \begin{align*} \left(\sum_{k=1}^{n+1} a_k\right)^2&=\left(\sum_{k=1}^n a_k\right)^2+2a_{n+1}\sum_{k=1}^n a_k+a_{k + 1}^2\\ &\ge\sum_{k=1}^n a_k^3 + 2a_{n + 1} \sum_{k=1}^n a_k + a_{n + 1}^2. \end{align*} Now it only needs $$2\sum_{k=1}^n a_k+a_{n+1}\ge a_{n + 1}^2.$$

Since for any $i,j\in\{0,1,2\cdots,n\},j>i$, then there is $$a_j-a_i\le j-i,$$ we have $$a_1\ge a_{n+1}-n,\ a_2\ge a_{n+1}-(n-1),\ \cdots$$ so $$2\sum_{k=1}^n a_k+a_{n+1}\ge (2n+1)a_{n+1}-(n+1)n$$ $$\Longleftrightarrow (a_{n+1}-n)(a_{n+1}-n-1)\le 0.$$ From here I can't work out. Thank you everyone.

Take $a_1 = \cdots = a_n$. This can't be right. Or is it $a_j - a_i \ge j-i$ instead? — , Jul 06 '13 at 17:39
If you use induction, you should first show that the statement is true in the base case. — dreamer, Jul 06 '13 at 18:11
The fact that there is equality for $a_i = i$ makes me think that, combined with the hypotheses implying that $a_i \le i$, that equality would be somehow used in the proof. — marty cohen, Jul 06 '13 at 18:37
http://math.stackexchange.com/questions/61798/generalisation-of-the-identity-sum-limits-k-1n-k3-bigg-sum-limits-k — jimjim, Jul 07 '13 at 00:56
http://math.stackexchange.com/questions/61482/proving-the-identity-sum-limits-k-1n-k3-large-sum-limits-k-1n-k — jimjim, Jul 07 '13 at 00:58
http://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct — jimjim, Jul 07 '13 at 01:00

zuggg · Accepted Answer · 2013-07-07T04:34:32.833

We can solve this using induction, like you suggested. The case $n=1$ is quite easy: $(a_1)^2\geq(a_1)^3$ since $0\leq a_1\leq 1$. Now assume that for some $n > 0$, the property is true for $n-1$. Let $S_n=\sum_{k=0}^n a_k=\sum_{k=1}^n a_k$. \begin{align*} (S_n)^2-\sum_{k=1}^n (a_k)^3 &= (a_n)^2 + 2a_nS_{n-1} + (S_{n-1})^2 - \sum_{k=1}^n (a_k)^3 \\ &\geq (a_n)^2 + 2a_nS_{n-1}-(a_n)^3 \\ &\geq a_n \left( -(a_n)^2 + a_n + 2S_{n-1}\right) . \end{align*} Since $a_n\geq 0$, we only need to prove the positivity of $$ -(a_n)^2 + a_n + 2S_{n-1} =: T_n. $$ Let $x=a_n-a_{n-1} \in [0,1]$. \begin{align*} T_n &= -(a_{n-1}+x)^2+a_{n-1}+x+2a_{n-1}+2S_{n-2}\\ &= -(a_{n-1})^2+a_{n-1}+2S_{n-2}-2xa_{n-1}+2a_{n-1}-x^2+x\\ &= T_{n-1} + x(1-x) + 2a_{n-1}(1-x)\\ &\geq T_{n-1} \end{align*} Since $T_1=-(a_1)^2+a_1 \geq 0$, another simple induction proves that $T_n\geq 0$, which concludes this proof.

How to prove that $\left(\sum\limits_{k=1}^{n}a_{k}\right)^2\ge\sum\limits_{k=1}^{n}a_k^3$?

1 Answers1