已知a>0,b>0,c>0,abc=1,证明1/a^3(b+c)+1/b^3(a+c)+c^3(a+b)>=3/2

由于1/a^3(b+c)=abc/a^2(ab+bc)=1/a^2(1/b+1/c)令x=1/a,y=1/b,z=1/c,又由于abc=1，a、b、c∈R+，有xyz=1,且x、y、z∈R+，于是只需证明x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥3/2.因为x^2/(y+z)+(y+z)/4≥x,y^2/(x+z)+(x+z)/4≥y,z^2/(x+y)+(x+y)/4≥z,以上三式相加得x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥(x+y+z)/2≥3(xyz)^(1/3)/2=3/2。得证
基本不等式

“c^3(a+b)”你肯定只有这些吗