for any $x$ and for any $t>1$ prove this inequality: $$ g(t,x)=(t+1)^x+(t-1)^x-2t^x>0. $$ I know that for $x=0$ and $x=1$ we have $g(t,0)=g(t,1)=0$. For the first integers like $t=2$ the solution is $x>1/2$, because the derivate of the function $g(2,x)$ is positive in $x=1$. How can I prove that the solution is $x>1$ for any $t>1$?
Thanks
$f(y)=y^x$ is a convex function for $x<0$ and $x>1$ so for every $y_1<y_2$ and $s\in [0,1]$ you have that
$f(s y_1+(1-s)y_2)<sf(y_1)+(1-s)f(y_2)$
but if you consider
$y_1:=t-1$
$y_2:=t+1$
$s=\frac{1}{2}$
you have that
$(\frac{1}{2}(t-1)+\frac{1}{2}(t+1))^x<\frac{1}{2}(t-1)^x+\frac{1}{2}(t+1)^x$
and so
$t^x<\frac{1}{2}[(t-1)^x+(t+1)^x]$
$2t^x<(t-1)^x+(t+1)^x$
For $x=0$ and $x=1$ does not true the inequality.
For $0<x<1$ the function $f(y)=y^x$ changed concavity and so it is true the inequality
$(t-1)^x+(t+1)^x<2t^x$