# How prove this $\left(\sum_{k=1}^{n}a_{k}\right)^2\ge\sum_{k=1}^{n}(a_k)^3$

let $a_{n}\ge a_{n-1}\ge\cdots\ge a_{0}= 0$,and for any $i,j\in\{0,1,2\dots,n\},j>i$,then have $$a_{j}-a_{i}\le j-i$$ show that $$\left(\sum_{k=1}^n a_k \right)^2\ge\sum_{k=1}^n (a_k)^3$$

my idea is by in Mathematical induction

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

when $n=k+1$ $$\left(\sum_{i=1}^{k+1} a_i\right)^2=\left(\sum_{i=1}^k a_i\right)^2+2a_{k+1}\sum_{i=1}^k a_i+(a_{k + 1})^2\ge\sum_{i=1}^k (a_i)^3 + 2a_{k+1} \sum_{i=1}^k a_i + (a_{k + 1})^2$$ $$\Longleftrightarrow 2\sum_{i=1}^k a_i+a_{k+1}\ge (a_{k + 1})^2$$ since for any $i,j\in\{0,1,2\cdots,n\},j>i$,then have $$a_j-a_i\le j-i$$ we have $$a_1\ge a_{k+1}-k,a_2\ge a_{k+1}-(k-1),\cdots$$ so $$2\sum_{i=1}^k a_i+a_{k+1}\ge (2k+1)a_{k+1}-(k+1)k$$ $$\Longleftrightarrow (a_{k+1}-k)(a_{k+1}-k-1)\le 0$$ following I can't work.Thank you everyone can help,Thank you everyone.

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Take $a_1 = \cdots = a_n$. This can't be right. Or is it $a_j - a_i \ge j-i$ instead? – user27126 Jul 6 '13 at 17:39
Surely, something is missing, probably $a_0=0$ ? – zuggg Jul 6 '13 at 17:49
@zuggg,Thank you, I have edit – math110 Jul 6 '13 at 17:58
@Sanchez,No, That's right – math110 Jul 6 '13 at 17:59
If you use induction, you should first show that the statement is true in the base case. – dreamer Jul 6 '13 at 18:11

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.