Is it true that if $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n≥x^n+y^n$ for $n>1$? Let $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n \ge x^n+y^n$ where $n=2$.


Is it true that if $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n \ge x^n+y^n$ for $n \ge 1$?


I have just been checked it is true with $1< a, b, x, y \le 1000$ with $n=3,4$ by my computer. I am looking for a proof if this is true inequality.
 A: Let
$d_n
= a^n+b^n
$
and let $a+b=d$.
$\begin{array}\\
d_n(a+b)
&=(a^n+b^n)(a+b)\\
&=a^{n+1}+ba^n+ab^n+b^{n+1}\\
&=a^{n+1}+b^{n+1}+ab(a^{n-1}+b^{n-1})\\
&=d_{n+1}+abd_{n-1}\\
\text{so}\\
d_{n+1}
&=d_n(a+b)-abd_{n-1}\\
&=dd_n-abd_{n-1}\\
\end{array}
$
Similarly,
if
$e_n=x^n+y^n
$
and
$e=x+y$,
$e_{n+1}
=ee_n-xye_{n-1}
$.
$d_1 \ge e_1$
and
$\begin{array}\\
d_2
&=a^2+b^2\\
&=a^2+2ab+b^2-2ab\\
&=(a+b)^2-2ab\\
&\ge(x+y)^2-2xy\\
&=e_2\\
\end{array}
$
If
$d_n \ge e_n$
and
$d_{n-1} \ge e_{n-1}
$,
then
$\begin{array}\\
d_{n+1}
&=dd_n-abd_{n-1}\\
&\ge ee_n-xyd_{n-1}
\qquad d\ge e, xy\ge ab, d_n\ge e_n\\
&= ee_n-xy(e_{n-1}-e_{n-1}+d_{n-1})\\
&= ee_n-xye_{n-1}-xy(e_{n-1}-d_{n-1})\\
&= e_{n+1}+xy(d_{n-1}-e_{n-1})\\
&\ge e_{n+1}
\qquad d_{n-1}\ge e_{n-1}\\
\end{array}
$
Therefore
$d_n \ge e_n$
for all $n \ge 1$.
A: Assuming that the inequality holds for $n - 1$ and $n - 2$, we can use induction.
\begin{align}
a^n + b^n &= a^{n - 1}(a + b) - a^{n - 1}b + b^{n - 1}(a + b) - ab^{n - 1} \\
&=(a + b)(a^{n - 1} + b^{n - 1}) - ab(a^{n - 2} + b^{n - 2}) \\
&\geq (x + y)(x^{n - 1} + y^{n - 1}) - xy(a^{n - 2} + b^{n - 2}) \\
&= x^n + y^n + xy(x^{n - 2} + y^{n - 2} - a^{n - 2} - b^{n - 2}) \\
&> x^n + y^n
\end{align}
Indeed, when $n = 1$ we are given the inequality. When $n = 2$, we have
\begin{align}
a^2 + b^2 &= (a + b)^2 - 2ab \\
&\geq (x + y)^2 - 2xy \\
&=x^2 + y^2.
\end{align}
Let me know if there are any mistakes.
