大师作文网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;
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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;
x、y、z是三角形三边长,
求证：(x+y-z)(y+z-x)(z+x-y)≤xyz;
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
(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
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