求助一道数学题a^2*b/2c+b^2*c/2d+c^2*d/2a+d^2*a/2b≥ab+cd
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求助一道数学题a^2*b/2c+b^2*c/2d+c^2*d/2a+d^2*a/2b≥ab+cd
貌似是用一个类似均值不等式的不等式解法,把这几项拆成两组,分别用均值可以得出,但是我现在忘记怎么拆了,
貌似是用一个类似均值不等式的不等式解法,把这几项拆成两组,分别用均值可以得出,但是我现在忘记怎么拆了,
a^2*b/2c+b^2*c/2d+c^2*d/2a+d^2*a/2b≥ab+cd
证明:∵a^2*b/2c+b^2*c/2d≥2√(a^2*b^3/4d) =ab√(b/d)
且c^2*d/2a+d^2*a/2b≥2√(c^2*d^3/4b)=cd√(d/b)
∴a^2*b/2c+b^2*c/2d+c^2*d/2a+d^2*a/2b
≥ab√(b/d)+cd√(d/b)
≥ab+cd
证明:∵a^2*b/2c+b^2*c/2d≥2√(a^2*b^3/4d) =ab√(b/d)
且c^2*d/2a+d^2*a/2b≥2√(c^2*d^3/4b)=cd√(d/b)
∴a^2*b/2c+b^2*c/2d+c^2*d/2a+d^2*a/2b
≥ab√(b/d)+cd√(d/b)
≥ab+cd
求助一道数学题a^2*b/2c+b^2*c/2d+c^2*d/2a+d^2*a/2b≥ab+cd
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