Graphically and numerically it is obvious but I'm looking for an analytical reasoning. Just maximizing the left hand side does not yield an analytical expression for the maximum. I also tried some known bounds for $\log$, but all of them had "overshoot".


2 Answers 2


Put $y=\frac{1}{\sqrt{x}}$ to get $(y+\frac{1}{y})\log 2-2\log y \geq 0, y>1$. Differentiating gives $\frac{1}{y}((y-\frac{1}{y})\log 2 -2)$, so we have critical points when $(y-\frac{1}{y})\log 2=2$. Solving the quadratic in $y$ and taking the positive root gives $y=\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$. It suffices to check the inequality for this value and the end points. ($y \to 1$ and $y \to +\infty$).

Put $y=1$ to get $2\log 2-2 \log(1)$ which is clearly non-negative. If the inequality holds for $y=\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$, then since the gradient $\frac{1}{y}((y-\frac{1}{y})\log 2 -2)$ is clearly non-negative when $y>\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$, the inequality holds as $y \to +\infty$.

Finally we prove the inequality when $y=\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$. Rewrite the inequality as $2y\log 2 \geq (y-\frac{1}{y})\log 2+2\log y$ and substitute in the value to get $2(1+\sqrt{(\log 2)^2+1}) \geq 2+2\log(\frac{1+\sqrt{(\log 2)^2+1}}{\log 2})$. Put $z=\sqrt{(\log 2)^2+1}$, so we get $z+\log(\log 2) \geq \log (1+z)$, or equivalently, $e^z\log 2 \geq 1+z$. The function $e^x\log 2-(1+x)$ is an increasing function for $x \geq 1$ (The gradient is $e^x\log 2-1 \geq e\log 2-1>0$), and $z>\sqrt{1+0.69^2}>1.2$, so it suffices to show that $e^{1.2}\log 2 \geq 2.2$. Note that $\log 2>0.69$ and $e^{1.2}>1+1.2+\frac{1.2^2}{2}+\frac{1.2^3}{6}=3.208>3.2$, and $3.2(0.69)=2.208>2.2$, so we are done.

  • $\begingroup$ Nice! I did not expect that a direct method could have worked. But with the coordinate transformation it turns out to be feasible after all. Thanks a lot. $\endgroup$ Mar 13, 2013 at 8:25

when you take the derivative you will get such an expression

$$x^{-\frac{\sqrt{x}}{1+x}} \left(-\frac{1}{\sqrt{x} (1+x)}+\left(\frac{\sqrt{x}}{(1+x)^2}-\frac{1}{2 \sqrt{x} (1+x)}\right) \text{Log}[x]\right)$$

Then solving for

$$\left(-\frac{1}{\sqrt{x} (1+x)}+\left(\frac{\sqrt{x}}{(1+x)^2}-\frac{1}{2 \sqrt{x} (1+x)}\right) \text{Log}[x]\right)=0$$

will give you some roots. one of them is $0.0907763$ and the other one is $11.0161$. If one checks carefully then it can be seen that the root $0.0907763$ yields the maximum which is $1.94011$.

Okay more analytical:

From the equation above we get:

$\log(x)=\frac{\frac{1}{\sqrt{x} (1+x)}}{\left(\frac{\sqrt{x}}{(1+x)^2}-\frac{1}{2 \sqrt{x} (1+x)}\right)}$

If we insert this into


and do some algebra we get


where $-1<y<\infty$ and originally $y=\frac{1+x}{x-1}$

Now it is easy to see that this function has a maxima at $-1$ which is less than $\log 2$

  • $\begingroup$ Aren't there too many decimal places in this analytical answer? :) $\endgroup$
    – J.H.
    Mar 13, 2013 at 0:09
  • $\begingroup$ The solution way is analytic the roots are not. I am not using any algorithm to search the maximum of the given function. $\endgroup$ Mar 13, 2013 at 0:14
  • $\begingroup$ @SeyhmusGüngören I think the OP wants to prove the statement using only pen and paper. Technically, if we're allowed to compute those roots using a computer, why not just ask it to graph the inequality/confirm that it is true? $\endgroup$
    – L. F.
    Mar 13, 2013 at 0:21
  • $\begingroup$ okay. Just a second. I will edit. $\endgroup$ Mar 13, 2013 at 0:24
  • $\begingroup$ How did you get $$x^{-\frac{\sqrt x}{1+x}}$$ in that derivative?? $\endgroup$
    – DonAntonio
    Mar 13, 2013 at 0:57

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