erdos不等式✍️

全称为Erdos－Mordell（鄂尔多斯市—门德尔）不等式，简称E-M不等式。

内容介绍

设P是ΔABC内任意一点，P到ΔABC三边BC,CA,AB的距离分别为PD=p,PE=q,PF=r，记PA=x,PB=y,PC=z。则

x+y+z≥2*(p+q+r)

证法介绍1

因为P,E,A,F四点共圆，PA为直径，则有:EF=PA*sinA。

在ΔPEF中，据余弦定理得:

EF^2=q^2+r^2-2*q*r*cos(π-A)=q^2+r^2-2*q*r*cos(B+C)

=(q*sinC+r*sinB)^2+(q*cosC-r*cosB)^2≥(q*sinC+r*sinB)^2，

所以有 PA*sinA≥q*sinC+r*sinB，即

PA=x≥q*(sinC/sinA)+r*(sinB/sinA) (1)。

同理可得:

PB=y≥r*(sinA/sinB)+p*(sinC/sinB) (2)，

PC=z≥p*(sinB/sinC)+q*(新浪/sinC) (3)。

(1)+(2)+(3)得:

x+y+z≥p*(sinB/sinC+sinC/sinB)+q*(simC/sinA+sinA/sinC)+r*(sinA/sinB+sinB/sinA)≥2*(p+q+r)。命题成立。

证法介绍2

设∠BP=2α，∠CPA=2β,∠APB=2γ，令它们内角平分线分别为:t1,t2,t3。则只需证明更强的不等式

x+y+z≥2*(t1+t2+t3)。

事实上，注意到内角平分线公式有:

t1=(2*y*z*cosα)/(y+z)≤(√y*z)*cosα，

同理可得: t2≤(√z*x)*cosβ,t3≤(√x*y)*cosγ。

由于α+β+γ=π，所以由嵌入不等式可得:

2*(t1+t2+t3)≤2*(√y*z)*cosα+2*(√z*x)*cosβ+2*(√x*y)*cosγ≤x+y+z。证毕。

证法介绍3

The proof of the inequality is based on the following

先给出一个引理

Lemma

引理

For the quantities x, y, z, p, q, r in ΔABC, we have ax ≥ br + cq, by ≥ ar + cp, and cz ≥ aq + bp.

在ΔABC中，对数值 x, y, z, p, q, r，恒有 ax ≥ br + cq, by ≥ ar + cp, cz ≥ aq + bp.

Proof of Lemma

下证引理成立：

For the proof we construct a trapezoid as shown. The diagram makes the first inequality ax ≥ br + cqobvious. The other two are shown similarly.

(That we do have a trapezoid follows from counting the angles at vertex A: they do sum up to 180°.)

由三角形两边之和大于第三边即可证引理成立。

The Erdös-Mordell Inequality

If O is a point within a triangle ABC whose distances to the vertices are x, y, and z, then

x + y + z ≥ 2(p + q + r).

回到原待证不等式。

Proof

证明：

From the lemma we have ax ≥ br + cq, by ≥ ar + cp, and cz ≥ aq + bp. Adding these three inequalities yields

x + y + z ≥ (b/a + a/b)r + (c/a + a/c)q + (c/b + b/c)p.

由引理得 x + y + z ≥ (b/a + a/b)r + (c/a + a/c)q + (c/b + b/c)p.

But the arithmetic mean-geometric mean inequality insures that the coefficients of p, q, and r are each at least 2, from which the desired result follows.

由均值不等式（AM-GM不等式）得p,q,r的系数 ≥ 2。

故待证不等式得证。

Observe that the three inequalities in the lemma are equalities if and only if O is the circumcenter of ΔABC, for in this case the trapezoids become rectangles.

观察引理中三个不等式取等号时当且仅当O是ΔABC的外心（此时梯形变成长方形）。

参考资料