Balkan 2012 shortlist

Let $m_0$ be an integer at which $f$ achieves its maximum. Then $h(n)\ge f(m_0)g(n-m_0)$ for all $n\in \mathbb{Z}$ and $$\sum _{n\in \mathbb{Z}}h(n)\ge f(m_0)\sum _{n\in \mathbb{Z}}g(n-m_0)=f(m_0)\sum _{n\in \mathbb{Z}}g(n).$$ Similarly, if $n_0$ is an integer at which $g$ achieves its maximum, then $$\sum _{n\in \mathbb{Z}}h(n)\ge g(n_0)\sum _{n\in \mathbb{Z}}f(n-n_0)=g(n_0)\sum _{n\in \mathbb{Z}}f(n).$$ Combining these yields $$\sum _{n\in \mathbb{Z}}h(n)\ge (f(m_0){)}^{1/q}{\left(\sum _{n\in \mathbb{Z}}g(n)\right)}^{1/q}(g(n_0){)}^{1/p}{\left(\sum _{n\in \mathbb{Z}}f(n)\right)}^{1/p}.$$ Since $$(f(m_0){)}^{1/q}{\left(\sum _{n\in \mathbb{Z}}f(n)\right)}^{1/p}={\left((f(m_0){)}^{p-1}\sum _{n\in \mathbb{Z}}f(n)\right)}^{1/p}\ge {\left(\sum _{n\in \mathbb{Z}}(f(n){)}^p\right)}^{1/p},$$ and $$(g(n_0){)}^{1/p}{\left(\sum _{n\in \mathbb{Z}}g(n)\right)}^{1/q}\ge {\left(\sum _{n\in \mathbb{Z}}(g(n){)}^q\right)}^{1/q},$$ the conclusion follows.

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Alternative solution.

We apply H\"older's inequality to obtain $${\left(\sum _{k\in \mathbb{Z}}(f(k){)}^{q(p-1)}\right)}^{1/q}{\left(\sum _{k\in \mathbb{Z}}(g(n-k){)}^{p(q-1)}\right)}^{1/p}\ge \sum _{k\in \mathbb{Z}}(f(k){)}^{p-1}(g(n-k){)}^{q-1}.$$ Hence $$h(n){\left(\sum _{k\in \mathbb{Z}}(f(k){)}^p\right)}^{1-1/p}{\left(\sum _{k\in \mathbb{Z}}(g(k){)}^q\right)}^{1-1/q}\ge \sum _{k\in \mathbb{Z}}(f(k){)}^p(g(n-k){)}^q.$$ Summing over $n$ gives $$\left(\sum _{n\in \mathbb{Z}}h(n)\right){\left(\sum _{k\in \mathbb{Z}}(f(k){)}^p\right)}^{1-1/p}{\left(\sum _{k\in \mathbb{Z}}(g(k){)}^q\right)}^{1-1/q}\ge \left(\sum _{k^{\prime }\in \mathbb{Z}}(f(k^{\prime }){)}^p\right)\left(\sum _{k^{\prime }\in \mathbb{Z}}(g(k^{\prime }){)}^q\right).$$ The conclusion follows.