x_1+x_2>A型不等式证明题,证明函数的自变量x在题设条件下的两处取值之和大于常数A,形式上推陈出新.例1已知函数f(x)=ln x-mx~2,g(x)=1/2 mx~2+x,m∈R,令F(x)=f(x)+g(x),若m=-2,正实数x_1,x_2满足F(x_1)+F(x_2)+x_1x_2=0,求证:x_1x_2≥(5~(1/2)-1)/2.证明当m=-2时,F(x)=lnx+x~2+x,则由F(x_1)+F(x_2)+x_1x_2=0,可得lnx_1x_2+(x_1+x_2)+(x_1+x_2)~2-x_1x_2=0, x_1 + x_2> type A inequalities prove that the argument x of the function x is greater than the constant A in the two conditions under the assumption of the set conditions, and formally evolve. Example 1 Known function f (x) = ln x-mx Let f (x) = f (x) + g (x) be a positive real number x_1, x_2, g (x) = 1 / 2mx ~ 2 + (x_1) + F (x_2) + x_1x_2 = 0, and prove that: x_1x_2≥ (5 ~ (1/2) -1) / 2. It is proved that F (x) = lnx + x ~ 2 + x, then x (x_1 + x_2) ~ 2-x_1x_2 = 0 can be obtained from F (x_1) + F (x_2) + x_1x_2 =