2026.6.9 闲话

第一人称合集 .

歌：Gimme Love - サブサキ feat. 重音テト & 音街ウナ .

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对数和不等式 (Log Sum Inequality)

对于长度为 \(n\) 的非负实数序列 \(\{a\}, \{b\}\)，

\[\sum_{i=1}^na_i\ln\dfrac{a_i}{b_i}\ge\left(\sum_{i=1}^na_i\right)\ln\dfrac{\sum_{i=1}^na_i}{\sum_{i=1}^nb_i} \]

A-side

在 \(C=\dfrac{\sum_{i=1}^na_i}{\sum_{i=1}^nb_i}\) 处用切线放缩 \(\ln x\ge1+\ln C-\dfrac Cx\)：

\[\begin{aligned}\sum_{i=1}^na_i\ln\dfrac{a_i}{b_i}&\ge\left(1+\ln\dfrac{\sum_{i=1}^na_i}{\sum_{i=1}^nb_i}\right)\sum_{i=1}^na_i-\dfrac{\sum_{i=1}^na_i}{\sum_{i=1}^nb_i}\sum_{i=1}^nb_i\\&=\left(\sum_{i=1}^na_i\right)\ln\dfrac{\sum_{i=1}^na_i}{\sum_{i=1}^nb_i}\end{aligned} \]

B-side

\(\lambda_i=\dfrac{b_i}{\sum_{i=1}^nb_i},\,t_i=\dfrac{a_i}{b_i},\,f(t)=t\ln t\)，Jensen 不等式：

\[\begin{aligned}\sum_{i=1}^n\lambda_if(t_i)&\ge f\left(\sum_{i=1}^n\lambda_it_i\right)\\\sum_{i=1}^n\dfrac{b_i}{\sum_{k=1}^nb_k}\dfrac{a_i}{b_i}\ln\dfrac{a_i}{b_i}&\ge\dfrac{\sum_{i=1}^na_i}{\sum_{i=1}^nb_i}\ln \dfrac{\sum_{i=1}^na_i}{\sum_{i=1}^nb_i}\\\sum_{i=1}^na_i\ln\dfrac{a_i}{b_i}&\ge\left(\sum_{i=1}^na_i\right)\ln\dfrac{\sum_{i=1}^na_i}{\sum_{i=1}^nb_i}\end{aligned} \]

这是可以解决 2025.4.5 闲话中的问题的 .

话说，到底告诉了我们什么呢、