<s>[INST] 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.
 [/INST] 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.
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