已知a>b>c,求证a²b+b²c+c²a>ab²+bc²+ca²
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已知a>b>c,求证a²b+b²c+c²a>ab²+bc²+ca²
已知a>b>c,求证a²b+b²c+c²a>ab²+bc²+ca²
已知a>b>c,求证a²b+b²c+c²a>ab²+bc²+ca²
由a > b > c,有a-b > 0,b-c > 0,a-c > 0.
故(a-b)(b-c)(a-c) > 0,展开即得a²b+b²c+c²a > ab²+bc²+ca².