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）。