4

Let $a,b,c$ and $d$ be real numbers with $a+d = b+c$, prove that $(a-b)(c-d)+(a-c)(b-d)+(d-a)(b-c) \geq 0$.

Should I substitute in the given condition for $a$ and $b$ and see if things simplify? Or should I use the arithmetic-geometric mean inequality?

Jacob Willis
  • 1,601
  • Have you tried anything? It seems that using substitution from the condition is a reasonable place to start. – DMcMor Dec 23 '15 at 22:46

2 Answers2

5

Note that $(a−b)(a−d)+(c−d)(b−d)+(d−a)(b−c)=2 (a-b) (c-d)=2(c-d)^2 \geq 0$

math635
  • 1,595
2

Write $a=x+y,b=x+z,c=x-z,d=x-y$ for $x,y,z\in\Bbb R$ (exercise: why is this possible?). Then your inequality is $$(y-z)(y-z)+(y+z)(y+z)+(-2y)(2z)\geq 0$$ After expanding you should be able to finish this.

Wojowu
  • 26,600