Question

数学竞赛不等式问题x、y、z是三角形三边长,求证：(x+y-z)(y+z-x)(z+x-y)≤xyz;

Answer 1

证明：

(x+y-z)(y+z-x) = [y+(x-z)][y-(x-z)]=y^2-(x-z)^2≤y^2 (1)

(x+y-z)(z+x-y) = [x+(y-z)][x-(y-z)]=x^2-(y-z)^2≤x^2 (2)

(y+z-x)(z+x-y)= [z+(y-x)][z-(y-x)]=z^2-(y-x)^2≤z^2 (3)

(1),(2),(3)相乘得

[(x+y-z)(y+z-x)(z+x-y)]^2≤(xyz)^2,

所以(x+y-z)(y+z-x)(z+x-y)≤xyz