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: $$xt^{x-y}y^{\frac{1}{x}} \geq xt^{x-y}$$


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

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

Thanks in advance !

  • 1
    $\begingroup$ I haven't used Jensen's inequality much, but I still have a sense that you might be able to use ithere... $\endgroup$ Aug 4, 2022 at 19:09

1 Answer 1


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 $t = b/a$.

  • $\begingroup$ i was scratching my head for a long time trying to figure out is it 0 and not 1 . Thanks for the rectification ^^ $\endgroup$ Aug 4, 2022 at 19:14
  • 1
    $\begingroup$ 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. $\endgroup$ Aug 4, 2022 at 20:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .