从7#开始 五改三 挑战纯几何 证明正三形 20140328

isee

基于楼上，有一纯几何证明，并是证明其加强，也就是大名鼎鼎的几何不等式：

厄尔多斯-蒙代尔 ( Erdos-Mordell ) 不等式：三角形内一点到三边距离之和小于等于到三顶点距离之和的一半

也有推广式forum.php?mod=viewthread&tid=2478，$n=3$时，便是Erdos-Mordell 不等式。



==
果然，不简单

isee

基于楼上，有一纯几何证明，并是证明其加强，也就是莫名的几何不等式：

厄尔多斯-蒙代尔 ( Erdos-Mordell  ...
isee 发表于 2014-3-28 16:52

而这个，从1935年提出，已经有各种漂亮的证明。

因此，这题并是其等号成立的情形，于是，纯几何法有了……

isee

而最典型几何证明，有如下两种：
en.wikipedia.org/wiki/Erdős–Mordell_inequality
http://zh.wikipedia.org/wiki/埃尔德什－莫德尔不等式

可视化证明：forumgeom.fau.edu/FG2007volume7/FG200711.pdf

isee

随便点了一上，forumgeom.fau.edu/FG2014volume14/FG201411.pdf，这费马点的推广吧？还是复数证明

forumgeom.fau.edu 不错啊，意外的惊喜


对，人教的海盗呢？看到英文突然想到他了

hbghlyj

Zhiqui Lu
Theorem. Let $p, q, r \geq 0$ and let $\alpha+\beta+\gamma=\pi$. Then we have the inequality
\[\tag1
p+q+r \geq 2 \sqrt{q r} \cos \alpha+2 \sqrt{r p} \cos \beta+2 \sqrt{p q} \cos \gamma .
\]
Proof. We consider the following quadratic function of $x$ :
\[\tag2
x^2-2(\sqrt{r} \cos \beta+\sqrt{q} \cos \gamma) x+q+r-2 \sqrt{q r} \cos \alpha .
\]
Then a quarter of the discriminant is
\[
\frac{1}{4} \Delta=(\sqrt{r} \cos \beta+\sqrt{q} \cos \gamma)^2-(q+r-2 \sqrt{q r} \cos \alpha) .
\]
Since $\alpha+\beta+\gamma=\pi$, we have
\[
\cos \alpha=-\cos (\beta+\gamma)=-\cos \beta \cos \gamma+\sin \beta \sin \gamma
\]
Using the above identity, the discriminant can be simplified as
\[
\Delta=-(\sqrt{r} \sin \beta-\sqrt{q} \sin \gamma)^2 \leq 0 .
\]
Thus the expression (2) is always nonnegative. Letting $x=\sqrt{p}$, we get (1).
Corollary. Let $x^{\prime}, y^{\prime}, z^{\prime}$ be the length of the angle bisectors of $\angle B P C, \angle C P A$, and $\angle A P B$, respectively. Then we have
\[
p+q+r \geq 2\left(x^{\prime}+y^{\prime}+z^{\prime}\right) .
\]
Proof. We have
\[
\begin{aligned}
x^{\prime} &=\frac{2 q r}{q+r} \cos \gamma \leq \sqrt{q r} \cos \gamma \\
y^{\prime} &=\frac{2 p r}{p+r} \cos \beta \leq \sqrt{p r} \cos \beta, \\
z^{\prime} &=\frac{2 p q}{p+q} \cos \alpha \leq \sqrt{p q} \cos \alpha
\end{aligned}
\]
The corollary follows from the theorem.
Remark. Since $x^{\prime} \geq x, y^{\prime} \geq y$ and $z^{\prime} \geq z$, the corollary implies the Erdös-Mordell inequality
\[
p+q+r \geq 2(x+y+z)
\]

hbghlyj

Zhiqin Lu
The Math Club
University of California, Irvine
November 28, 2007
...
slide 34
The key observation
We let $\vec{A}, \vec{B}, \vec{C}$ be vectors in $R^3$ with length $a, b, c$, respectively. Assume that the angle between $\vec{A}, \vec{B}, \vec{C}$ are $\pi / 2+\alpha / 2, \pi / 2+\beta / 2, \pi / 2+\gamma / 2$, respectivley. Then we have
\[
\left(|\vec{A}|^2+|\vec{B}|^2+|\vec{C}|^2\right)^2 \geq 4\left(|\vec{A} \times \vec{B}|^2+|\vec{B} \times \vec{C}|^2+|\vec{C} \times \vec{A}|^2\right)
\]
For two matrices, we have $AB\ne BA$. Thus we define the commutator of the matrices as$$[A, B] = AB - BA$$
which measures the non-commutativity of the matrices.
Using the notion of commutator, we have the following result
Theorem
Let $A, B, C$ be $3 \times 3$ skew-symmetric matrices with zero diagonal parts. Then we have
$$\left(\|A\|^2+\|B\|^2+\|C\|^2\right)^2 \geq 8\left(\|[A, B]\|^2+\|[B, C]\|^2+\|[C, A]\|^2\right)$$
The above inequality implies the Erdös-Mordell inequality.
Conjecture (Normal scalar curvature conjecture)
Let $A_1, \cdots, A_m$ be $n \times n$ symmetric matrices. Then we have
\[
\left(\sum\left\|A_i\right\|^2\right)^2 \geq 2 \sum_{i<j}\left\|\left[A_i, A_j\right]\right\|^2 .
\]
Conjecture (Böttcher-Wenzel Conjecture)
Let $A, B$ be two $n \times n$ matrices. Then
\[
2\|[A, B]\|^2 \leq\left(\|A\|^2+\|B\|^2\right)^2 .
\]
In summer of 2007, I proved both conjectures.
I made the following conjecture:
Conjecture (Zhiqin Lu)
Let $A, B$ be the bounded trace class operators in a separable Hilbert space. Then we have
\[
2\|[A, B]\|^2 \leq\left(\|A\|^2+\|B\|^2\right)^2,
\]
where the normal is defined as
\[
\|A\|=\sqrt{\operatorname{Tr}\left(A^* A\right)} .
\]