Inequality involving an exponent

Question

I wish to prove the following inequality $$x^{\frac{3}{x-1}} > 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}, \quad x > 1.$$ Graphically the above inequality appears to be true since if one plots $$g(x) = x^{\frac{3}{x-1}} - \left (1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} \right )$$ it appears as though $g(x) > 0$ for all $x > 1$. I have however been unable to prove analytically this is true.

I know $$1 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} < 4 \quad \mbox{for all} \,\, x> 1$$ and $$1 < x^{\frac{3}{x - 1}} < \mathrm{e}^3 \quad \mbox{for all} \,\, x > 1,$$ but neither of these bounds seem to help me very much.

Any pointers in the right direction would be greatly appreciated.

Answer 1

Hint :

Consider the Taylor expansion at infinity of your exponent :

$$x^{\frac{3}{x-1}} = x ^{3\ \big(\frac{1}{x} +\frac{1}{x^2} + \frac{1}{x^3} +...\big)}$$

Answer 2

Hint: Taking logs, and using $u>\ln (1+u), u> 0,$ we see it suffices to show

$$\ln x \ge (x-1)/3\cdot (1/x + 1/x^2 + 1/x^3), \ \ x> 1.$$

Both sides are $0$ at $x=1,$ so it suffices to show the inequality for the derivatives of each side.

Answer 3

We want to prove that $$ \forall x\in(0,1),\qquad x^{\frac{3x}{x-1}}>1+x+x^2+x^3\tag{1}$$ but it is enough to prove that: $$ \forall x\in(0,1),\qquad 3x\log(x)<(x-1)(x+x^2+x^3)=x^4-x\tag{2} $$ or: $$ \forall x\in(0,1),\qquad 1+3\log(x) < x^3\tag{3} $$ or (by replacing $x$ with $z^{1/3}$): $$ \forall z\in(0,1),\qquad 1+\log(z) < z \tag{4} $$ that is trivial by the concavity of the $\log$ function.