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A math problem concerned about monotonicity


Given that f(x) is a monotonously increasing function and g(x)=f(x)-f(1-x). Now if for certain real numbers x_1, x_2, g(x_1)+g(x_2)>0, compare x_1+x_2 and 1.

Here we will prove that $latex  x_1+x_2>1$.

Obviously, when x_1>=1-x_1, x_2>=1-x_2 while the equalities don’t hold simultaneously, we can easily deduce the condition; at the same time we can get x_1+x_2>1.

While if x_1<=1-x_1 and x_2<=1-x_2, apparently we have

f(x_1)<=f(1-x_1), f(x_2)<=f(1-x_2), so f(x_1)+f(x_2)<=f(1-x_1)+f(1-x_2), so g(x_1)+g(x_2)<=0.

Now consider if x_1>=1-x_1 and x_2<=1-x_2. Thus we have x_1>=1/2 and x_2<=1/2. So at the same time we have
x_2<=x_1 and x_2<=1-x_2

Now if x_1+x_2<=1, then x_1<=1-x_2, then x_2<=1-x_1.

So we have x_2<=1-x_1<=x_1<=1-x_2

So f(x_2)<=f(1-x_1)<=f(x_1)<=f(1-x_2)

So f(x_2)+f(x_1)<=f(1-x_1)+f(1-x_2)

Which indicates that the condition holds only if x_1+x_2>1.

Symmetrically we can prove the same when x_1<=1-x_1 and x_2>=1-x_2.

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