Prove inequality $y^{(n-1)}+x^n\leq x^{(n-1)}+y^n$

Question

Is there a short way to show that $$y^{n-1}+x^n\leq x^{n-1}+y^n$$ for $n\in\mathbb N$ and $1<x<y$?

I have tried to show it with starting from the left side without rewriting the equation $$y^{n-1}+x^{n}=y^{n-1}+x\cdot x^{n-1}\leq x\cdot y^{n-1}+x^{n-1}\leq y\cdot y^{n-1} +x ^{n-1}.$$ But I am not sure how to show the second to last step without rewriting the equation.

Answer 1

$f(x)=x^{n}-x^{n-1}$ is an increasing function on $(1,\infty)$ because $f'(x)=nx^{n-1}-(n-1)x^{n-2} >0$ ( since $x >1 >1-\frac 1 n$). The required inequality simply says $x<y$ implies $f(x) <f(y)$.

Answer 2

The inequality is equivalent to $$ x^{n-1}(x-1)\leq y^{n-1}(y-1) $$ This inequality again follows directly from the condition $1<x<y$ since all terms are positive (this is where you need the condition $x,y>1$).

Answer 3

Need to show:

$x^n -x^{n-1} \lt y^n-y^{n-1}$;

$x^{n-1}(x-1) \lt y^{n-1}(y-1)$.

Since $1<x<y$ :

$(y-1)>(x-1)>0$, and

$y^{n-1} > x^{n-1}>0$(why?).

Answer 4

$$1<x<y\implies 0<x-1<y-1.$$

Multiply the left inequalities $n-1$ times and the right ones once to get

$$x^{n-1}(x-1)<y^{n-1}(y-1).$$

Equality is not possible.