Prove inequality using M.V.T.

Question

For $x>2$, prove that $(x-1)e^{\tfrac{2}{x}}-(x-2)e^{\tfrac{1}{x}}<e$ using the mean value theorem.

I can't find the function to applying the mean value theorem.

Give some advice or hint. Thank you!

For any fixed $x>2$, I tried to using M.V.T on $[1/x,2/x]$ to the function et1−t.

But, i have a trouble in using the relation $1/x<c_{x}<2/x$, where $c_{x}$ derived from M.V.T..

Answer 1

Let $f(x)=(x-1)e^{\frac{2}{x}}-(x-2)e^{\frac{1}{x}}.$

We'll prove that $f$ decreases for $x>2$.

Indeed, $$f'(x)=\frac{e^{\frac{1}{x}}}{x^2}\left((x^2-2x+2)e^{\frac{1}{x}}-(x^2-x+2)\right)$$ and it's enough to prove that $g(x)>0,$ where $$g(x)=\ln(x^2-x+2)-\frac{1}{x}-\ln(x^2-2x+2).$$ We have: $$g'(x)=\frac{2x-1}{x^2-x+2}+\frac{1}{x^2}-\frac{2x-2}{x^2-2x+2}=\frac{(2-x)(3x^2-2x+2)}{x^2(x^2-x+2)(x^2-2x+2)}<0$$ for $x>2,$ which says $$g(x)>\lim_{x\rightarrow+\infty}g(x)=0.$$ Id est, $$f(x)<\lim_{x\rightarrow2^+}f(x)=e$$ and we are done!

Answer 2

Just a hint:

Think of the function $f(a) = (x-a)e^{\frac{a}{x}}$; its derivative wrt $a$ is $-\frac{a}{x}e^{\frac{a}{x}}$

Then, there exists a $c \in [1,2]$ such that $(x-1)e^{\tfrac{2}{x}}-(x-2)e^{\tfrac{1}{x}} \ge (x-1)e^{\tfrac{1}{x}}-(x-2)e^{\tfrac{2}{x}} = \frac{c}{x}e^{\frac{c}{x}} $

What can be the max value of RHS?