Prove that: $x^x-y^y\ge(x-y)^x-(x-y)^y$, where $x\ge y>0.$

Question

I want to analyze the following inequality:

Let $x,y\in\Bbb R^{+}$, then prove that: $$x^x-y^y\ge(x-y)^x-(x-y)^y$$ holds for all $x\ge y$.

I observed that, for possible simplification, dividing both side of the inequality into any of the expressions like $x^x, x^y, y^y, y^x$ doesn't work.

Using the identity $x^x=e^{x\ln x}$ and $y^y=e^{y\ln y}$ doesn't help.

I used the well-known inequality $e^x≥x+1$ and obtained:

$$x^x=e^{x\ln x}≥x\ln x+1$$

or

$$y^y=e^{y\ln y}≥y\ln y+1$$

But, this also doesn't help.

Some observations:

Big problem occurs when, $x=0.002, y=0.001$. $$-0.0055≈x^x-y^y\ge(x-y)^x-(x-y)^y≈-0.0068$$

If $x=5,y=4.5$ then: $$(x-y)^x-(x-y)^y≈-0.001<0$$

If $x=5,y=3$ then: $$(x-y)^x-(x-y)^y≈24>0$$

If $x-y=1$, then: $(x-y)^x-(x-y)^y=0$. Thus

$$x^x-(x-1)^{x-1}\ge 0$$

which is correct, since $x>1$.

If $x=y>0$, then the equality trivially holds: $0≥0$

The possible inequality plot is as follows:

I also include the reverse inequality plot:

$$x^x-y^y<(x-y)^x-(x-y)^y$$

How can we analyze this inequality? This doesn't seem possible without using calculus.

Maybe the inequality is simple. However, I could not find a good point, unfortunately. Also, I'm having trouble understanding the first graph that Wolfram Alpha presents.

Answer 1

The inequality is false in general. The results in Reza Rajaei's answer show that the inequality holds when $y\geq \frac{1}{e}$.

But conversely, when $y\lt \frac{1}{e}$ there will always be an $x$ making the inequality false. To see why, consider the ratios $f_1(x)=\frac{(x-y)^x-(x-y)^y}{x-y}$ and $f_2(x)=\frac{x^x-y^y}{x-y}$ and let us look at their limit when $x$ tends to $y$ from above.

If we put $\phi(x)=x^x$, then $f_2(x)=\frac{\phi(x)-\phi(y)}{x-y}$ tends to $l_2=\phi'(y)=(1+\log(y))y^y$ when $x\to y$ from above.

Next, we can write

$$ f_1(x) = \frac{(x-y)^x-(x-y)^y}{x-y} \\ = \frac{(x-y)^y}{x-y}\big((x-y)^{x-y}-1\big) \sim_{x\to y} \frac{(x-y)^y}{x-y}\log\big((x-y)^{x-y}\big) $$

because $u-1\sim \log(u)$ when $u\to 1$. It follows that $f_1(x)\sim_{x\to y} (x-y)^y \log\big(x-y\big)$, and hence that $f_1(x)\to l_1=0$ when $x\to y$ from above. Now, when $y\lt \frac{1}{e}$, we have $l_1\gt l_2$ and hence $f_1(x) \gt f_2(x)$ when $x$ is close enough to $y$.

When we look at a numerical example, say $y=\frac{1}{3}$, we see that $x$ needs to be VERY close to $y$ (the distance between $x$ and $y$ cannot exceed $10^{-7}$). This explains why the inequality seems true and first sight, and why Wolfram Alpha didn't see the counterexamples.

UPDATE With some more work, the following quantitative proof above can be "unrolled" into a quantitative proof with explicit constants, as explained below.

To construct our counterexempale, we choose a $y\lt\frac{1}{e}$, say $y=\frac{1}{3}$.

Above we noticed the qualitative result that $f_2(x)\to l_2$ when $x\to y$. We turn this into a quantitative result as follows. We apply a Taylor bound to $\phi$. Note that $\phi'(x)=(1+\log(x))\phi(x)$ and hence $\phi''(x)=\big(\frac{1}{x}+(1+\log(x))^2\big)\phi(x)$. Now, from $e^u\geq 1+u$ we deduce $e^2\geq 3$ and hence $\log(3)\leq 2$. For $x\in [y,1]$, we then have $|\phi(x)|\leq 1$ and

$$ |\phi''(x)|\leq \big(3+(1+\log(3))^2\big) \leq 3+(1+3)^2=12 \tag{1} $$

Using the Taylor remainder integral formula $\phi(x)-\phi(y)-\phi'(y)(x-y)=\int_{y}^{x}\phi''(t)\frac{(x-t)^2}{2}dt$, we deduce $|\phi(x)-\phi(y)-\phi'(y)(x-y)|\leq 12 \frac{(x-y)^3}{6}=2(x-y)^3$. Dividing by $x-y$, we obtain for $x\in [y,1]$,

$$ |f_2(x)-l_2| \leq 2(x-y)^2 \tag{2} $$

Similarly, we convert the limit $f_1(x)\to 0$ when $x\to y$ into a quantitative result. This part is more complicated because there are several substeps more. We have $f_1(x)=f_3(x)f_4(x)$, where

$$ f_3(x)=(x-y)^y\log(x-y), \ f_4(x)=\frac{(x-y)^{x-y}-1}{\log((x-y)^{x-y})} \tag{3} $$

Using the well-known inequality $(*):-1 \leq u\log(u)\leq 0$ for $u\in(0,1]$, we have (putting $u=(x-y)^{\frac{y}{2}}$)

$$ |f_3(x)| =\big|\frac{2}{y}(x-y)^y\log((x-y)^{\frac{y}{2}})\big|=\frac{2(x-y)^{\frac{y}{2}}}{y}|u\log(u)|\leq \frac{2(x-y)^{\frac{y}{2}}}{y}= 6 (x-y)^{\frac{1}{6}} \tag{4} $$

Similarly, using $(*)$ with $u=x-y$ we deduce $-1 \leq z\leq 0$ where $z=(x-y)\log(x-y)$ ; now the mean value theorem yields $|e^z-1|\leq |z|$, and hence $|(x-y)^{x-y}-1|\leq |(x-y)\log(x-y)|$. Using an argument similar to the proof of (4) above, we have $|(x-y)\log(x-y)|\leq 2\sqrt{x-y}$ and hence

$$ 1-2\sqrt{x-y} \leq (x-y)^{x-y} \leq 1 \tag{5} $$

So if we put $v=(x-y)^{x-y}$, the mean value theorem yields $ 1 \leq \frac{log(v)-\log(1)}{v-1} \leq \frac{1}{1-2\sqrt{x-y}}$ or $1-2\sqrt{x-y} \leq f_4(x) \leq 1$, and hence

$$ |f_4(x)| \leq 1+2\sqrt{x-y} \tag{6} $$

Multiplying (4) by (6), we deduce

$$ |f_1(x)|\leq 6 (x-y)^{\frac{1}{6}} (1+2\sqrt{x-y}) = 6(x-y)^{\frac{1}{6}}+12(x-y)^{\frac{2}{3}} \leq 18(x-y)^{\frac{1}{6}} \tag{7} $$

Combining (2) with (7), we obtain

$$ |f_1(x)-f_2(x)+l_2|\leq 2(x-y)^2+18(x-y)^{\frac{1}{6}} \leq 20(x-y)^{\frac{1}{6}} \tag{8} $$

Note that $|l_2|=(log(3)-1)\phi(\frac{1}{3})$, now

$$\log(3)=\int_{1}^{3}\frac{dt}{t}=\sum_{k=1}^8 \int_{1+\frac{k-1}{4}}^{1+\frac{k}{4}}\frac{dt}{t}\geq \sum_{k=1}^8\frac{1}{4+k}=\frac{28271}{27720} \tag{9}$$

and hence $\log(3)-1\geq\frac{551}{27720}\geq\frac{1}{100}$, on the other hand $\phi(\frac{1}{3})=\frac{1}{3}^{\frac{1}{3}} \geq \frac{1}{2}$ because $\frac{1}{3} \geq \frac{1}{8}$. Finally $|l_2|\geq \frac{1}{200}$, and we deduce from (8) that

$$ f_1(x)-f_2(x) \geq \frac{1}{200} - 20(x-y)^{\frac{1}{6}} \tag{10} $$

We see then that $f_1(x)\gt f_2(x)$ whenever $x-y\lt \frac{1}{4000^6}$ ; this is the case for example when $x-y=10^{-22}$.

Answer 2

There exists a partial solution which covers your examples, I think.

---

For $x\ge 1$, define $f(t)=x^t-(x-y)^t$, then $f'(t)=x^t\ln x-(x-y)^t\ln (x-y)\ge0$, which means $f(t)$ is an increasing function for $t\ge0$ and $x\ge1$. Hence:

$$f(x)\ge f(y)\implies x^x-(x-y)^x\ge x^y-(x-y)^y \ge y^y-(x-y)^y$$$$\implies x^x-y^y\ge(x-y)^x-(x-y)^y.$$

---

Another easy case is $x\ge 2y$. In this case $x-y\ge y$, and we have $(x-y)^y\ge y^y$ and $x^x\ge (x-y)^x.$

---

EDIT:

Another easy case is when $1\gt x\gt y\ge \frac {1}{e} $. In this case $0\lt x-y \lt1,$ so $(x-y)^x\le (x-y)^y$ because $f(t)=(x-y)^t$ is a decreasing function. Therefore, if $1\gt x\gt y\ge \frac {1}{e} $; $$x^x-y^y\ge0 \ge (x-y)^x- (x-y)^y; $$

because $x^x$ is an increasing function when $x\ge \frac {1}{e}.$

---

Answer 3

Remark: I found that the desired inequality is false several hours before @Ewan Delanoy posted his answer. I was held up by something. I post my answer as well.

Consider the case $x > y$ and $y < 1$. Let $z = x - y > 0$. The desired inequality is written as $$(y + z)^{y+z} - y^y + z^y - z^{y+z} \ge 0$$ or $$\frac{(y + z)^{y+z} - y^y}{z} + \frac{1 - z^z}{z^{1 - y}} \ge 0.$$

For fixed $y\in (0, 1)$, by L'Hopital's rule, we have $$\lim_{z \to 0^{+}} \frac{1 - z^z}{z^{1 - y}} = 0$$ and $$\lim_{z \to 0^{+}} \frac{(y + z)^{y+z} - y^y}{z} = y^y(\ln y + 1).$$

Thus, if $\ln y + 1 < 0$, we have $$\lim_{z \to 0^{+}} \left(\frac{(y + z)^{y+z} - y^y}{z} + \frac{1 - z^z}{z^{1 - y}}\right) < 0.$$ Thus, the desired inequality is not true for some $z$ when $y < \mathrm{e}^{-1}$.