Popoviciu's inequality

hbghlyj

kuing 发表于 2021-7-21 08:13
易知 f(x) 是下凸函数，所以由 Popoviciu 不等式可知上式成立，即得证。

补充Popoviciu’s inequality的证明《不等式的秘密》第二卷—高级不等式 第六章
Example 1.17.1. If $f$ is a convex function

then
\begin{equation}\begin{aligned}&f(a) + f(b) + f(c) + f\left(a + b + c\over3\right)\\
≥ \frac43&\left(f\left(a + b\over2\right)+ f\left(b + c\over2\right)+ f\left(c + a\over2\right)\right)\end{aligned}\label1\end{equation}Solution. WLOG, suppose that $a ≥ b ≥ c$. Consider the following number sequences
\begin{aligned}(x) &= (a, a, a, b, t, t, t, b, b, c, c, c);\\
(y) &= (α, α, α, α, β, β, β, β, γ, γ, γ, γ);\end{aligned}where\begin{aligned}
t &={a + b + c\over3},\\
α &={a + b\over2},\\
β &={a + c\over2},\\
γ &={b + c\over2}.\end{aligned}Clearly, we have that $(y)$ is a monotonic sequence. Moreover\begin{gathered}
a ≥ α, 3a + b ≥ 4α, 3a + b + t ≥ 4α + 2β, 3a + b + 3t ≥ 4α + 3β,\\
3a + 2b + 3t ≥ 4α + 4β, 3a + 3b + 3t ≥ 4α + 4β + γ,\\
3a + 3b + 3t + c ≥ 4α + 4β + 2γ, 3a + 3b + 3t + 3c ≥ 4α + 4β + 4γ.\end{gathered}Thus $(x^∗) ≫ (y)$ and therefore $(x^∗) ≫ (y^∗)$. By Karamata inequality, we conclude
\[3 (f(x) + f(y) + f(z) + f(t)) ≥ 4 (f(α) + f(β) + f(γ)) ,\]
which is exactly the desired result. We are done.

hbghlyj

Wikipedia
Let $f$ be a function from an interval $ I\subseteq \mathbb {R} $ to $ \mathbb {R} $. If $f$ is convex, then for any three points $x, y, z$ in $I$,\begin{equation}{\frac {f(x)+f(y)+f(z)}{3}}+f\left({\frac {x+y+z}{3}}\right)\geq {\frac {2}{3}}\left[f\left({\frac {x+y}{2}}\right)+f\left({\frac {y+z}{2}}\right)+f\left({\frac {z+x}{2}}\right)\right].\label2\end{equation}

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Darij Grinberg
Let $f$ be a convex function from an interval $I ⊆ ℝ$ to $ℝ$, let $x_1, x_2, x_3$ be three points from $I$, and let $w_1, w_2, w_3$ be three nonnegative reals such that $w_2 + w_3\ne0, w_3 + w_1\ne0$ and $w_1 + w_2\ne0$. Then,
\begin{equation}\label3\begin{array}{l}w_{1} f\left(x_{1}\right)+w_{2} f\left(x_{2}\right)+w_{3} f\left(x_{3}\right)+\left(w_{1}+w_{2}+w_{3}\right) f\left(\frac{w_{1} x_{1}+w_{2} x_{2}+w_{3} x_{3}}{w_{1}+w_{2}+w_{3}}\right) \\ \geq\left(w_{2}+w_{3}\right) f\left(\frac{w_{2} x_{2}+w_{3} x_{3}}{w_{2}+w_{3}}\right)+\left(w_{3}+w_{1}\right) f\left(\frac{w_{3} x_{3}+w_{1} x_{1}}{w_{3}+w_{1}}\right)+\left(w_{1}+w_{2}\right) f\left(\frac{w_{1} x_{1}+w_{2} x_{2}}{w_{1}+w_{2}}\right)\end{array}\end{equation}

hbghlyj

\eqref{2}包含在\eqref{3}里
但\eqref{1}不包含在\eqref{3}里
能否把\eqref{1}和\eqref{3}统一起来

kuing

hbghlyj 发表于 2023-5-8 03:55
\eqref{2}包含在\eqref{3}里
但\eqref{1}不包含在\eqref{3}里
能否把\eqref{1}和\eqref{3}统一起来

Popoviciu 当然是 (2)，它的加权推广是 (3)。

至于 (1) 它虽然也成立，但它弱于 (2)，理由：由 (2) 的
\[\frac{f(x)+f(y)+f(z)}3+f\left(\frac{x+y+z}3\right)\geqslant\frac23\left[f\left(\frac{x+y}2\right)+f\left(\frac{y+z}2\right)+f\left(\frac{z+x}2\right)\right],\]
再加上
\[\frac{f(x)+f(y)}3\geqslant\frac23f\left(\frac{x+y}2\right),\frac{f(y)+f(z)}3\geqslant\frac23f\left(\frac{y+z}2\right),\frac{f(z)+f(x)}3\geqslant\frac23f\left(\frac{z+x}2\right),\]
就得到 (1)，所以 (1) 是弱的结论，意义不大。