$k\in\Bbb N$, for each $n\in \Bbb N$ we let $f(n)$ be the number of solutions $(x_1,\cdots,x_k)\in \Bbb Z^k$ of the inequality $$|x_1|+\cdots +|x_k|\le n$$ that is $$f(n)= card\left(\{ (x_1,\cdots,x_k)\in \Bbb Z^k: |x_1|+\cdots +|x_k|\le n\}\right)$$ Then prove that: $$f(n-1)f(n+1)\le [f(n)]^2$$
My intention is to compute first, $$g(n)= card\left(\{ (x_1,\cdots,x_k)\in \Bbb Z^k: |x_1|+\cdots +|x_k|= n\}\right)$$
since, $$\color{red}{f(n) = f(n-1)+g(n)}$$
But I am wondering whether $$\color{red}{g(n)= 2*card\left(\{ (x_1,\cdots,x_k)\in \Bbb N^k: x_1+\cdots +x_k= n\}\right)}$$
I have tried to use the induction I couldn't succeed.