Proving or disproving: If $0<a<b<1$, then $(1-a)^b>(1-b)^a$

Question

Prove or disprove

If $0<a<b<1$, then $$(1-a)^b>(1-b)^a$$

I think this looks true when evaluating the differential equation $\frac{dy}{dx}=-y$ with initial condition $y(0)=1$ using euler method

As an example with step size a=0.2 and b=0.3

with step size $a=0.2$

Evaluate the value of $y$, 3 times by Euler method with step size a=0.2 will get

$y_{a3}=1(1-0.2)^3=1(1-0.2)^{10(0.3)}$

While using step size of $0.3$ and repeat 2 times

$y_{b2}=1(1-0.3)^2=1(1-0.3)^{10(0.2)}$

Plotting this in (x,y) axis, one can compare at $x=0.6 ,y_{a3}>y_{b2}$. I tried with different value of a and b, the inequality looks true as saying Euler method with smaller step size will be the upper boundary for bigger step size for this differential equation. Which also mean the exact solution of this exponential function is the uppest boundary for all step size bigger than dx.

I tried binomial expansion, but it only makes it much complicated.

$(1-a)^b=1-ab+\frac{b(b-1)}{2!}(-a)^2+..$

$(1-b)^a=1-ab+\frac{a(a-1)}{2!}(-b)^2+..$

I can only show for the 3rd term

$\frac{b(b-1)}{2!}(-a)^2>\frac{a(a-1)}{2!}(-b)^2$

since $b>a$

I do not found any way to proof nth term of $(1-a)^b$ will always bigger than $(1-b)^a$, this is where i stuck.

Answer 1

Let $f(x)=\frac{\ln(1-x)}{x}.$

Thus, we need to prove that $f$ decreases, which is not so obvious without derivative.

We have: $$f'(x)=-\frac{\ln(1-x)+\frac{x}{1-x}}{x^2}.$$ Now, let $g(x)=\ln(1-x)+\frac{x}{1-x}.$

Thus, $$g'(x)=\frac{1}{(1-x)^2}-\frac{1}{1-x}=\frac{x}{(1-x)^2}>0,$$ which says $$g(x)>g(0)=0$$ and we got that $f$ decreases.