要用一下下麵一題

First (by + cz)(bz + cy) <= 2(b^2y^2 + c^2z^2) can be proved:

(by + cz)(bz + cy) = yz(b^2 + c^2) + bc (y^2 + z^2). (b^2y^2 + c^2z^2) - yz(b^2 + c^2) = b^2y(y-z) + c^2z(z-y) = (y-z)(b^2y-c^2z) >= 0, similarly (b^2y^2 + c^2z^2) - bc (y^2 + z^2) >= 0.

We have similar inequalities for (cz+ax)(cx+az) etc.

Let U = a^2x^2, V = b^2y^2, W = c^2z^2, we want to prove:

U/(V+W) + V/(W+U) + W/(U+V) >= 3/2, which is the other problem. (which can be proved by adding 3 to both sides, then both sides multiply 2, rewrite R=U+V etc, change to problem (R+S+T)*(1/R+1/S+1/T) >= 9.