Let $f(x)$ be monotone decreasing on $(0,+\infty)$, such that $$0<f(x)<\lvert f'(x) \rvert,\qquad\forall x\in (0,+\infty).$$

Show that $$xf(x)>\dfrac{1}{x}f\left(\dfrac{1}{x}\right),\qquad\forall x\in(0,1).$$

My ideas:

Since $f(x)$ is monotone decreasing, $f'(x)<0$, hence $$f(x)+f'(x)<0.$$ Let $$F(x)=e^xf(x)\Longrightarrow F'(x)=e^x(f(x)+f'(x))<0$$ so $F(x)$ is also monotone decreasing. Since $0<x<1$, $$F(x)>F\left(\dfrac{1}{x}\right)$$ so $$e^xf(x)>e^{\frac{1}{x}}f\left(\frac1x\right).$$ So we must prove $$e^{\frac{1}{x}-x}>\dfrac{1}{x^2},\qquad0<x<1$$ $$\Longleftrightarrow \ln{x}-x>\ln{\dfrac{1}{x}}-\dfrac{1}{x},0<x<1$$ because $0<x<1,\dfrac{1}{x}>1$ so I can't. But I don't know whether this inequality is true. I tried Wolfram Alpha but it didn't tell me anything definitive.

PS: This problem is from a Chinese analysis problem book by Huimin Xie. enter image description here

  • $\begingroup$ Note that $xe^{x}$ is an increasing function if $x\in\left(0,1\right)$ so $xe^{x}<\frac{1}{x}e^{1/x}$ and then your inequality is false (the inequality in the comment). $\endgroup$ – Marco Cantarini Dec 23 '14 at 13:57
  • $\begingroup$ oh,this inequality is not true,that mean my idea is not usefull? $\endgroup$ – china math Dec 23 '14 at 14:01
  • $\begingroup$ maybe this books problem is not true? maybe we can take countexapmle? $\endgroup$ – china math Dec 23 '14 at 14:05
  • $\begingroup$ @MarcoCantarini, you are maybe wrong. See wolframalpha.com/input/…. $\endgroup$ – Alex Silva Dec 23 '14 at 14:05
  • $\begingroup$ To clearify the comment of @MarcoCantarini: $x\mapsto xe^x$ is increasing for all $x>0$. Thus $xe^x<1/xe^{1/x}$ for $x\in(0,1)$. Isnt, however, the function one should study $x\mapsto xe^{1/x}$? $\endgroup$ – mickep Dec 23 '14 at 14:07

A tentative of proof:

Put $g(x)=x^2f(x)-f(1/x)$. We have

$$g^{\prime}(x)=2xf(x)+x^2f^{\prime}(x)+\frac{1}{x^2}f^{\prime}(\frac{1}{x})=A+B+C$$ with $\displaystyle A=x^2(f(x)+f^{\prime}(x))$, $\displaystyle B=\frac{1}{x^2}(f(\frac{1}{x})+f^{\prime}(\frac{1}{x}))$, and $\displaystyle C=x(2-x)f(x)-\frac{1}{x^2}f(\frac{1}{x})$.

You have proved that $A<0$ and $B<0$. We have $$C =x(2-x)f(x)-\frac{1}{x^2}f(\frac{1}{x})=-(x-1)^2 f(x)+\frac{1}{x^2} g(x)$$ Hence $\displaystyle g^{\prime}(x)-\frac{1}{x^2}g(x)<0$. Put $h(x)=g(x)\exp(1/x)$, we have $\displaystyle h^{\prime}(x)=(g^{\prime}(x)-\frac{1}{x^2}g(x))\exp(1/x)$, and hence $h$ is decreasing. As $g(1)=0$, we have $h(1)=0$, $h(x)>0$ for $x\in (0,1)$, and we are done.

  • 1
    $\begingroup$ wa! Very very Nice! Thank you+1 $\endgroup$ – china math Dec 23 '14 at 14:29

I think the answer by @Kelenner is really good. This answer is just to prove the inequality $$ e^{1/x-x}>1/x^2,\quad 0<x<1.\tag{*} $$ that was discussed in the post/comments. We apply the logarithm, and since the logarithm is monotonically increasing, the inequality $(*)$ is equivalent to $$ 2\ln x>x-\frac{1}{x},\quad 0<x<1.\tag{**} $$ Let $$ g(x)=2\ln x-x+\frac{1}{x}. $$ Then $g(1)=0$ and a differentiation (and simplification) gives $$ g'(x)=-\frac{(x-1)^2}{x^2}. $$ Hence $g'(x)<0$ for $0<x<1$ (so $g$ is monotonically decreasing) and it follows that $g(x)>0$ for $0<x<1$ and thus that $(**)$ holds. But $(**)$ was seen to be equivalent to $(*)$, and so the inequality $(*)$ is true.

Edit: I updated the solution without the change of variable, since I don't think it simplified anything in the end.

  • 1
    $\begingroup$ Nice!Thank you very much $\endgroup$ – china math Dec 23 '14 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.