Let \(0\lt r\lt 1\), \(a\ge 0\), and \(b\gt 0\). Writing \(s=r/(r-1)\lt 0\), one has \(ab\ge a^r/r+b^s/s\).

Let \(r\in \mathbb {R}\) and \(u:\{ 0,1\} ^n\to [0,\infty )\) have at least one positive value. Then \(\mathbb {E}[u^r]\gt 0\), with the convention \(0^0=1\) and \(0^r=0\) for \(r\ne 0\).

Let \(r\ne 0\), let \(u:\{ 0,1\} ^n\to [0,\infty )\), and assume \(A=\mathbb {E}[u^r]\gt 0\) under the uniform measure. Then \(\mathbb {E}[(u/A^{1/r})^r]=1\).

Let \(u,v:\{ 0,1\} ^n\to \mathbb {R}\) and nonzero \(A,B\in \mathbb {R}\), uniform expectation satisfies \(\mathbb {E}[uv]=AB\mathbb {E}[(u/A)(v/B)]\).

Let \(0\lt r\lt 1\) and \(u,v:\{ 0,1\} ^n\to [0,\infty )\). Then \(\mathbb {E}[uv]\ge M_r(u)M_{r/(r-1)}(v)\). For nonnegative functions \(u\), the extended mean is \(M_r(u)=(\mathbb {E}u^r)^{1/r}\) when \(r\gt 0\); the same formula applies for \(r\lt 0\) when \(u\) is positive everywhere. For positive \(u\), \(M_0(u)=\exp (\mathbb {E}\log u)\); if \(r\le 0\) and \(u\) has a zero, \(M_r(u)=0\). All expectations are uniform.

Let \(p\lt 1\), \(p\ne 0\), and \(f,g:\{ 0,1\} ^n\to [0,\infty )\). Then \(\mathbb {E}[fg]\ge M_p(f)M_{p/(p-1)}(g)\). For nonnegative functions \(u\), the extended mean is \(M_r(u)=(\mathbb {E}u^r)^{1/r}\) when \(r\gt 0\); the same formula applies for \(r\lt 0\) when \(u\) is positive everywhere. For positive \(u\), \(M_0(u)=\exp (\mathbb {E}\log u)\); if \(r\le 0\) and \(u\) has a zero, \(M_r(u)=0\). All expectations are uniform.