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# Understand the problem

Given $a_1\ge 1$ and $a_{k+1}\ge a_k+1$ for all $k\ge 1,2,\dots,n$, show that $a_1^3+a_2^3+\dots+a_n^3\ge (a_1+a_2+\dots+a_n)^2$
Inequalities
Easy
##### Suggested Book
Inequalities by BJ Venkatachala

Do you really need a hint? Try it first!

Use induction.
Given the inequality for $n=k$, the inequality for $n=k+1$ follows if $a_{k+1}^2-a_{k+1}-2(a_1+a_2+\cdots +a_k)\ge 0$.
Note that, as $a_1\ge 1$, the hypothesis implies that $a_n\ge 1$ for all $n\in\mathbb{N}$.
From hint 3, $a_{k+1}^2-a_{k+1}-2(a_1+a_2+\cdots +a_k)=a_{k+1}(a_{k+1}-1)-2(a_1+a_2+\cdots +a_k)\ge a_{k}(a_{k}+1)-2(a_1+a_2+\cdots +a_k)=a_{k}^2+a_{k}-2(a_1+a_2+\cdots +a_k)=a_{k}^2-a_{k}-2(a_1+a_2+\cdots +a_{k-1})$. Repeating this procedure, we eventually arrive at $a_{k+1}^2-a_{k+1}-2(a_1+a_2+\cdots +a_k)\ge a_1^2-a_1\ge 0$.    QED

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