设a,b,c是正实数,求证:(1/2a)+(1/2b)+(1/2c)≥[1/(b+c)]+[1/(c+a)]+[1/(a+b)]
热心网友
1/2a+1/2b+1/2c=(1/4a+1/4b)+(1/4b+1/4c)+(1/4a+1/4c)可以先证明1/4a+1/4b≥1/(a+b)......,1/4b+1/4c≥1/(b+c)........1/4a+1/4c≥1/(a+c)........,再间接证明1/2a+1/2b+1/2c≥1/(a+b)+1/(b+c)+1/(a+c)成立下面来证明:要证明1/4a+1/4b≥1/(a+b)即证明(a+b)/4ab≥1/(a+b),即证明(a+b)^2≥4ab,即证明a^2+b^2+2ab≥4ab即证明a^2-2ab+b^2≥0,即证明(a-b)^2≥0,显然成立所以1/4a+1/4b≥1/(a+b)同理可以证明1/4b+1/4c≥1/(b+c),1/4a+1/4c≥1/(a+c)所以原式成立