柯西不等式的证明 1/(2a)+1/(2b)+1/(2c)>=1/(a+b)+1/(b+c)+1/(c+a)

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柯西不等式的证明 1/(2a)+1/(2b)+1/(2c)>=1/(a+b)+1/(b+c)+1/(c+a)
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柯西不等式的证明 1/(2a)+1/(2b)+1/(2c)>=1/(a+b)+1/(b+c)+1/(c+a)
柯西不等式的证明 1/(2a)+1/(2b)+1/(2c)>=1/(a+b)+1/(b+c)+1/(c+a)

柯西不等式的证明 1/(2a)+1/(2b)+1/(2c)>=1/(a+b)+1/(b+c)+1/(c+a)
用均值不等式可知:
1/a+1/b=(a+b)/(ab)≥4/(a+b).(1)
1/a+1/c=(a+c)/(ac)≥4/(a+c).(2)
1/c+1/b=(c+b)/(cb)≥4/(c+b).(3)
(1)+(2)+(3)然后左右两边同时除以4得:
1/(2a)+1/(2b)+1/(2c)≥1/(b+c)+1/(c+a)+1/(a+b).

(a+b)/(ab)>=4(a+b)/(a+b)^2=4/(a+b)