先证a^3+b^3≥a^2b+b^2a,由排序不等式,这是显然的,即1/(a^3+b^3+abc)≤1/(a^2b+b^2a+abc)=1/ab(a+b+c)同理,1/(b^3+c^3+abc)≤1/bc(a+b+c)1/(a^3+c^3+abc)≤1/ac(a+b+c)三式子相加,1/(a^3+b^3+abc)+1/(b^3+c^3+abc)+1...