# Prove the following lemma : $\frac{a^x}{b^y}\ge \frac{xa^{x-y}-yb^{x-y}}{x-y}$

I've seen this Lemma on an AOPS topic: https://artofproblemsolving.com/community/c6h2136817p15662370

$$\forall \hspace{2mm}a,b,x,y \geq 0$$ (with $$x\neq y$$)

we have:

$$\frac{a^x}{b^y}\ge \frac{xa^{x-y}-yb^{x-y}}{x-y}$$

I've been able to prove it when $$x > y$$

But not when $$x < y$$

If anyone can prove that this lemma is absolutely true or communicate a link i would be really grateful.

Here's my proof for the case $$x> y$$ (correct me if i'm wrong).

First note that we can divide by $$b^{x-y}$$ on both sides , Hence the inequality is equivalent to: $$\frac{a^x}{b^y.b^{x-y}} \geq \frac{x\frac{a^{x-y}}{b^{x-y}}-y}{x-y}$$ or: $$\frac{a^x}{b^x} \geq \frac{x\frac{a^{x-y}}{b^{x-y}}-y}{x-y}$$

Notice that we can replace $$\frac{a}{b}$$ by a variable t , Then the inequality to prove turns into this: $$t^x \geq \frac{xt^{x-y}-y}{x-y}$$ or: $$t^x+\frac{y}{x-y}\geq \frac{xt^{x-y}}{x-y}$$

Since $$x-y >0 :$$ $$(x-y)t^x+y\geq xt^{x-y}$$

Now by weighted AM-GM: $$(x-y)t^x+y \geq x (t^{x(x-y)}y)^{\frac{1}{x}}$$

or: $$(x-y)t^x+y \geq xt^{x-y}y^{\frac{1}{x}}$$

Now it remains to proof that: $$xt^{x-y}y^{\frac{1}{x}} \geq xt^{x-y}$$

or:

$$y^{\frac{1}{x}} \geq 1$$ $$\Longleftrightarrow$$ $$y \geq 1$$

I would like to know where i am failing , any hints are greatly appreciated.

Define $$f(t) = \frac{xt^y-yt^x}{x-y}$$, the original problem is equivalent to $$f(t)\leq 1$$. You can see $$f'(1)=0$$ is the only local extreme point. and it can be verified $$f''(1)<0$$ so that $$f(1)=1$$ is the maximum value, where $$t = b/a$$.
• Nope, I was wrong. I see it now. You don't even need to consider $f''(t).$ You can just say: $f'(t)>0$ for $0<t<1,$ $f'(1)=0$, and $f'(t) < 0$ for $t>1.$ Anyway this is a clever answer because one automatically starts trying functions where $x$ and $y$ are the variables, but the fact you don't have to do this is cool. Aug 4, 2022 at 20:34