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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 $x_2>=1-x_2$.