已知a,b,c∈R,求证(a²/b)+(b²/c)+(c²/a)≥a+b+c已知a,b,c∈R*,求证(a²/b)+(b²/c)+(c²/a)≥a+b+c
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![已知a,b,c∈R,求证(a²/b)+(b²/c)+(c²/a)≥a+b+c已知a,b,c∈R*,求证(a²/b)+(b²/c)+(c²/a)≥a+b+c](/uploads/image/z/5208009-33-9.jpg?t=%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%E2%88%88R%2C%E6%B1%82%E8%AF%81%28a%26%23178%3B%2Fb%29%2B%28b%26%23178%3B%2Fc%29%2B%28c%26%23178%3B%2Fa%29%E2%89%A5a%2Bb%2Bc%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%E2%88%88R%2A%2C%E6%B1%82%E8%AF%81%28a%26%23178%3B%2Fb%29%2B%28b%26%23178%3B%2Fc%29%2B%28c%26%23178%3B%2Fa%29%E2%89%A5a%2Bb%2Bc)
已知a,b,c∈R,求证(a²/b)+(b²/c)+(c²/a)≥a+b+c已知a,b,c∈R*,求证(a²/b)+(b²/c)+(c²/a)≥a+b+c
已知a,b,c∈R,求证(a²/b)+(b²/c)+(c²/a)≥a+b+c
已知a,b,c∈R*,求证(a²/b)+(b²/c)+(c²/a)≥a+b+c
已知a,b,c∈R,求证(a²/b)+(b²/c)+(c²/a)≥a+b+c已知a,b,c∈R*,求证(a²/b)+(b²/c)+(c²/a)≥a+b+c
运用均值不等式 b+a^2/b≥2a
c+b^2/c≥2b
a+c^2/a≥2c
两边相加易得