# Can we show such a reversed Cauchy-Schwarz for some convex functions?

Let's say $f$ is a convex function on $[0,+\infty]$ with $f(0)=f^{\prime}(0)=0$ and $\lim_{x \rightarrow +\infty} f(x)=+\infty$ ($f^{\prime}$ is stricty increasing),

Then I believe the following is true:

$$\lim_{x \rightarrow +\infty} \frac{\log[\int_{C}^{x} \frac{f^{\prime}(\tau)}{\tau} d\tau \int_{C}^{x} \tau f^{\prime} (\tau) d\tau]} {\log[f(x)]}=2$$ for any $C>0$

It is true that $\liminf \geq 2$ by Cauchy-Schwarz, but I am stuck at the $\limsup$ part. I can only show that $\limsup \leq 2$ when $\lim_{x \rightarrow +\infty} \frac{f(x)}{xf^{\prime}(x)}$ exists. However, this assumption seems not true in general (although I don't have a counterexample yet). I will greatly appreciate if anyone suggests a solution.

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For $f(x)=x^2$ the limit is equal to $+\infty$. –  Andrew May 20 '12 at 10:39
Should one read $f(x)^2$ in the denominator? But the RHS does not seem independent on $f$. –  Did May 20 '12 at 12:22
Thank both of you for pointing it out! Sorry for the typo. Careless last night. I've corrected with What with \log function. –  CABo May 20 '12 at 18:01