1

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 $k \geq 2$.

dmh
  • 2,958
  • 2
    Shouldn't the condition $x\ge y$ be there? – Asher2211 Sep 18 '21 at 18:04
  • @Asher2211 actually this is a point that confused me, I expected to have some condition on the relationship between $x$ and $y$, but the claim does not mention it. – dmh Sep 18 '21 at 18:07
  • At least for the cases of $n=2$ and $n=3$ we dont need such a condition. – dmh Sep 18 '21 at 18:09
  • Checking the errata in the text it mentions that we need the condition $x\geq y$ – dmh Sep 18 '21 at 18:10
  • 2
    Even for $n = 2$ you do need it. In fact for any $n$, if you take $x = 0, y > 0$ then the left hand side is $0$ but the right hand side is negative. – stochasticboy321 Sep 18 '21 at 18:14

1 Answers1

2

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*}

nejimban
  • 3,663