I will prove it by induction. It is true for $n=1$, so suppose it is true for $n$. We have: \[ \sum\limits_{i=1}^n a_i ^3 \geq \sum\limits_{i=1}^n a_i ^2 + 2\sum\limits_{i<j}^n a_ia_j \] Adding $a_{n+1}^3$ to both sides we have to prove that: \[ \sum\limits_{i=1}^n a_i ^2 + 2\sum\limits_{i<j}^{n} a_ia_j +a_{n+1}^3 \geq \sum\limits_{i=1}^{n+1} a_i ^2 + 2\sum\limits_{i<j}^{n+1} a_ia_j \] equivalently: \[ a_{n+1}^2+2a_{n+1}\sum\limits_{i=1}^n a_i \leq a_{n+1}^3 \] \[ a_{n+1}+2\sum\limits_{i=1}^n a_i \leq a_{n+1}^2 \] I will prove this by induction as well - suppose it is true for $n+1$, adding to both sides $a_{n+1}+a_{n+2}$ we are left to prove: \[ a_{n+1}^2 +a_{n+1}+a_{n+2} \leq a_{n+2}^2 \] \[ a_{n+1}(a_{n+1}+1)\leq a_{n+2}(a_{n+2}-1) \] which is true due to condition $a_{n+1}<a_{n+2}$ Hence the result

Valentin Vornicu wrote: Prove that for all positive integers $0<a_1<a_2<\cdots <a_n$ the following inequality holds: $(a_1+a_2+\cdots + a_n)^2 \leq a_1^3+a_2^3 + \cdots + a_n^3 $ Viorel Vajaitu Even simpler: $\displaystyle \left(\sum_{i=1}^n{a_i^3}\right)-\left(\sum_{i=1}^n{a_i}\right)^2=$ $\displaystyle \left(\sum_{i=1}^n{a_i^3-a_i^2}\right)-2\left(\sum_{i=2}^n{\left(a_i \sum_{j=1}^{i-1}{a_j}\right)}\right)$ $\displaystyle \ge \left(\sum_{i=1}^n{a_i^3-a_i^2}\right)-2\left(\sum_{i=2}^n{\left(a_i \sum_{j=1}^{i-1}{(a_i-j)}\right)}\right)$ $=\displaystyle \sum_{i=1}^n{a_i(a_i-i)(a_i-i+1)}$, that is a sum of non negative integers. []

An easy one! We can use induction and the facts that: 1) $a_n-a_i>=n-i$ 2)$a_n>=n$

Hmmm strange... the identity $\sum_{i=1}^n{i^3}=\left(\sum_{j=1}^n{j}\right)^2$ kills this problem without thought. A curious choice of a TST problem...