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KMTTQ, giả sử a=min{a,b,c}. Ta có: T=1b(a+b)+1c(b+c)+1a(c+a)−1a(b+c)−1b(c+a)−1c(a+b) =1a+b(1b−1c)+1b+c(1c−1a)+1c+a(1a−1b) Nếu b≤c thì T≥1c+a(1b−1c)+1c+a(1c−1a)+1c+a(1a−1b)=0. Nếu b>c thì T>1a+b(1b−1c)+1a+b(1c−1a)+1a+b(1a−1b)=0. Tóm lại: 1b(a+b)+1c(b+c)+1a(c+a)≥1a(b+c)+1b(c+a)+1c(a+b)≥92(ab+bc+ca).
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