<s>[INST] Show that $n(x-y)(xy)^{(n-1)/2} \leq x^n - y^n$ for nonnegative $x,y$ and integers $n$ Show for $x,y\geq 0$ and $n \in \mathbb N$ that:
$$n(x-y)(xy)^{(n-1)/2} \leq x^n - y^n$$
What I've tried
Proving by induction seems natural since the case for $n=2$ and $n=3$ are straightforward to show.
Assuming the claim is true for $n = k-1$, we want to show it for $n = k$:
$$(k-1)(x-y)(xy)^{(k-1-1)/2} \leq x^{(k-1)} - y^{(k-1)}$$
Now, multiplying both sides by $x+y$:
$$x^k - y^k - y^{k-1}x + x^{k-1}y \geq (k-1)(x+y)(x-y)(xy)^{(k-1-1)/2}$$
Recall that $(x+y)(x-y) = x^2 - y^2 \geq 2(x-y)\sqrt{xy}$
So then the whole LHS in the previous display is:
$$\geq 2k(x-y)(xy)^{(k-1)/2} - x^{k/2+1}y^{(k-1-1)/2} +x^{(k-1-1)/2}y^{n/2 + 1}$$
So if I could show that:
$$y^{k-1}x - x^{k-1}y - x^{k/2+1}y^{(k-1-1)/2} +x^{(k-1-1)/2}y^{k/2 + 1} \geq k(x-y)(xy)^{(k-1)/2}$$
I would be done. I think it should be possible to factor this expression or perhaps apply AM-GM in a clever way to show this, but I'm having trouble seeing it.
EDIT: as pointed out in the comments, we need the condition $x \geq y$. Notice on the LHS we have:
$$- y^{k-1}x + x^{k-1}y $$
but applying the condition on $x$ and $y$ we know that expression is at least $0$ since $-y^{k-1}x \geq -y^k$ and $x^{k-1}y \geq y^k$.
Now, applying AM-GM to $x+y$ we have that:
$$(k-1)(x+y)(x-y)(xy)^{(k-1-1)/2} \geq 2(k-1)(x-y)(xy)^{(k-1)/2}$$
$$= 2k(x-y)(xy)^{(k-1)/2} - 2(x-y)(xy)^{(k-1)/2}$$
But $2(x-y)(xy)^{(k-1)/2} \leq k(x-y)(xy)^{(k-1)/2}$ for all $k \geq 2$.
 [/INST] Yes, AM-GM. Let $x\ge y\ge0$, $n\in\mathbb N$. Then
\begin{align*}
x^n-y^n&=(x-y)\sum_{k=0}^{n-1}x^{n-1-k}y^k\\
&=\frac12(x-y)\sum_{k=0}^{n-1}\left(x^ky^{n-1-k}+x^{n-1-k}y^k\right)\\
&\ge\frac12(x-y)\sum_{k=0}^{n-1}2\sqrt{x^ky^{n-1-k}\cdot x^{n-1-k}y^k}\\
&=(x-y)\sum_{k=0}^{n-1}(xy)^{\frac{n-1}2}\\
&=n(x-y)(xy)^{\frac{n-1}2}.
\end{align*}
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