a,b,c,d>0 证(a/(b+2c+3d))+(b/(c+2d+3a))+(c/d+2a+3b))+(d/(a+2b

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a,b,c,d>0 证(a/(b+2c+3d))+(b/(c+2d+3a))+(c/d+2a+3b))+(d/(a+2b+3c))>=2/3

根据均值不等式：
ab≤(a²+b²)/2
ac≤(a²+c²)/2
ad≤(a²+d²)/2
bc≤(b²+c²)/2
bd≤(b²+d²)/2
cd≤(c²+d²)/2
以上六式相加：ab+ac+ad+bc+bd+cd≤3(a²+b²+c²+d²)/2
即：a²+b²+c²+d²≥(2/3)(ab+ac+ad+bc+bd+cd)
∴(a+b+c+d)²=a²+b²+c²+d²+2(ab+ac+ad+bc+bd+cd)≥(2/3+2)(ab+ac+ad+bc+bd+cd)
即：ab+ac+ad+bc+bd+cd≤(3/8)(a+b+c+d)²
于是a(b+2c+3d)+b(c+2d+3a)+c(d+2a+3b)+d(a+2b+3c)=4(ab+ac+ad+bc+bd+cd)
≤(3/2)(a+b+c+d)²
根据柯西不等式：
[a/(b+2c+3d)+b/(c+2d+3a)+c/(d+2a+3b)+d/(a+2b+3c)]·[a(b+2c+3d)+b(c+2d+3a)+c(d+2a+3b)+d(a+2b+3c)]≥(a+b+c+d)²
故(a+b+c+d)²≤[a/(b+2c+3d)+b/(c+2d+3a)+c/(d+2a+3b)+d/(a+2b+3c)]·[a(b+2c+3d)+b(c+2d+3a)+c(d+2a+3b)+d(a+2b+3c)]
≤[a/(b+2c+3d)+b/(c+2d+3a)+c/(d+2a+3b)+d/(a+2b+3c)]·[(3/2)(a+b+c+d)²]
即a/(b+2c+3d)+b/(c+2d+3a)+c/(d+2a+3b)+d/(a+2b+3c)≥2/3

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