Prove that: $x^x-y^y\ge(x-y)^x-(x-y)^y$, where $x\ge y>0.$ 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.
 A: 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 $x\ge \frac {1}{e}.$

A: 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 $y < \mathrm{e}^{-1}$.
