In the first proof, pay attention to A3+B3+C3-3abc = (A+B+C) (A2+B2+C2-AB-BC-CA), and get
3 ABC-[a2(b+ c-a)+B2(c+a-b)+C2(a+b-c)]
= a3+B3+C3-3 ABC+a(B2+C2-2bc)+b(C2+a2-2ca)+c(a2+B2-2ab)
=(a+b+c)(a2+B2+C2-a b-BC-ca)+a(B2+C2-2bc)+b(C2+a2-2ca)+c(a2+B2-2ab)
= 12(a+b+c)[(a-b)2+(b―c)2+(c―a)2]+a(b―c)2 ++ b(c―a)2+c(a―b)2
= 12(a+b-c)(a-b)2+ 12(b+c-a)(b―c)2+ 12(a+c―b)(c―a)2。
∵ A, B and C are three sides of a triangle, ∴ A+B-C > 0, B+C-A >; 0，a+c-b & gt； 0.
And (a-b) 2 ≥ 0, (b-c) 2 ≥ 0, (c-a) 2 ≥ 0, so the original inequality holds if and only if a=b=c, that is, △ABC is a regular triangle.
Example 2 It is known that A, B and C are positive numbers, which proves that:
( 1) AB+C+BC+A+CA+B ≥ 32。 (1963 Moscow Mathematical Olympics)
(2) a2b+c+b2c+a+c2a+b≥a+b+c2。 (The 2nd World Friendship Cup Mathematics Competition)
Proof (1)∫A b+ C+BC+A+CA+B-32
= 2a(a+b)(c+a)+2b(a+b)(b+c)+2c(b+c)(c+a)-3(a+b)(b+c)(c+a)2(a+b)(b+ c)(c+a)(c+a)
= 2(a3+B3+C3)-(a2 b+ab2+B2C+bc2+c2a+Ca2)2(a+b)(b+ c)(c+a)
= a3+B3-(a2 b+ab2)+B3+C3-(B2C+bc2)+B3+C3-(c2a+Ca2)2(a+b)(b+ c)(c+a)
=(a+b)(a-b)2+(b+c)(b-c)2+(c+a)(c-a)22(a+b)(b+c)(c+a)≥0，∴ab+c+bc+a +ca+b≥32。
(2) Using this identity, it is not difficult to prove that A2B+ C+B2C+A+C2A+B = (A+B+C) (AB+C+A+CA+B)-(A+B+C), the inequality ab+c+bc+a +ca+b≥32, A2B.
Example 3 If x, y and z are positive numbers, then y2-x2z+x+z2-y2x+y+x2-z2y+z ≥ 0. (Janus conjecture)
It is proved that let U = Y2-X2Z+X+Z2-Y2X+Y+X2-Z2Y+Z, V = Y2-Z2Z+X+Z2-X2X+Y+X2-Y2Y+Z,
Then u-v = z2-x2z+x+x2-y2x+y+y2-z2y+z = z-x+x-y+y-z = 0,
And u+v = (x2-y2) (1y+z-1z+x)+(y2-z2) (1z+x-1x+y)+(z2-x2) (/.
=(x2-y2)x-y(y+z)(z+x)+(y2-z2)y-z(z+x)(x+y)+(z2-x2)z-x(x+y)(y+z)
=(x+y)(x-y)2(y+z)(z+x)+(y+z)(y-z)2(z+x)(x+y)+(z+x)(z-x)2(x+y)(y+z)≥0，
Therefore, u = v>0. Therefore, y2-x2z+x+z2-y2x+y+x2-z2y+z ≥ 0.
Example 4 The positive real numbers X, Y and Z satisfy X, Y, z ≥ 1, which proves that X5-X2X5+Y2+Z2+Y5-Y2Y5+Z2+Z5-Z2Z5+X2+Y2 ≥ 0. (46th IMO test)
It is proved that xyz≥ 1, so
x5-x2x5+y2+z2≥x5-x2？ 6? 1xyzx5+(y2+z2)？ 6? 1xyz = x4-x2yzx4+yz (y2+z2) ≥ 2x4-x2 (y2+z2) 2x4+(y2+z2) 2 is also available.
y5-y2y 5+z2+x2≥2y 4-y2(z2+x2)2y 4+(z2+x2)2、z5-z2z 5+x2+y2≥2z 4-z2(x2+y2)2z 4+(x2+y2)2。
Let a=x2, b=y2 and c=z2, and the original inequality is simplified to prove 2a2-a (b+c) 2a2+(b+c) 2+2b2-b (c+a) 2b2+(c+a) 2+2c2-c (a+b) 2c2+(a).
8? 6a(a-b)+a(a-c)2 a2+(b+c)2+b(b-c)+b(b-a)2 B2+(c+a)2+c(c-a)+c(c-b)2c 2+(a+b)2≥0
8? 6∑cyc(a-b)( 12 a2+(b+c)2- 12 B2+(c+a)2)≥0？ 8? 6∑cyc(a-b)2(C2+c(a+b)+a2-a b+ B2(2 a2+(b+c)2)(2 B2+(c+a)2))≥0。