<s>[INST] How to prove that $\left(\sum\limits_{k=1}^{n}a_{k}\right)^2\ge\sum\limits_{k=1}^{n}a_k^3$? Let $$a_{n}\ge a_{n-1}\ge\cdots\ge a_{0}= 0,$$ and for any $i,j\in\{0,1,2\dots,n\},j>i$, 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.
 [/INST] 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.
</s>