全稱為Erdos－Mordell（鄂爾多斯—門德爾）不等式，簡稱E-M不等式。
基本介紹
- 中文名：鄂爾多斯—門德爾不等式
- 外文名：Erdos－Mordell Inequality
- 提出者：Erdos&Mordell
- 套用學科：數學
- 適用領域範圍：幾何不等式
內容,證法1,證法2,證法3（幾何法）,
內容
設P是ΔABC內任意一點，P到ΔABC三邊BC,CA,AB的距離分別為PD=p,PE=q,PF=r，記PA=x,PB=y,PC=z。則
x+y+z≥2*(p+q+r)
證法1
因為P,E,A,F四點共圓，PA為直徑，則有:EF=PA*sinA。
在ΔPEF中，據餘弦定理得:
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*(sinA/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)。命題成立。
證法2
設∠BP=2α，∠CPA=2β,∠APB=2γ，令它們內角平分線分別為:t1,t2,t3。則只需證明更強的不等式
x+y+z≥2*(t1+t2+t3)。
事實上，注意到內角平分線公式有:
t1=(2*y*z*cosα)/(y+z)≤(√y*z)*cosα，
同理可得: t2≤(√z*x)*cosβ,t3≤(√x*y)*cosγ。
由於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。證畢。
證法3（幾何法）
The proof of the inequality is based on the following
先給出一個引理
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，對數值 x, y, z, p, q, r，恆有 ax ≥ br + cq, by ≥ ar + cp, cz ≥ aq + bp.
Proof of Lemma
下證引理成立：
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°.)
由三角形兩邊之和大於第三邊即可證引理成立。
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).
回到原待證不等式。
Proof
證明：
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的係數 ≥ 2。
故待證不等式得證。
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.
觀察引理中三個不等式取等號時若且唯若O是ΔABC的外心（此時梯形變成長方形）。
