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Can someone explain why the geometric mean is less than the logarithmic mean? $$\sqrt{ab} \leq \frac{b-a}{\log b-\log a} $$

Jean Marie
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shabnam
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3 Answers3

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Let $\sqrt{ab}=:g$. Then $a=ge^{-p}$, $\>b=ge^p$ for some $p\geq0$, and we obtain $${b-a\over\log b-\log a}=g\>{2\sinh p\over 2p}\geq g\ ,$$ since $\sinh'(0)=1$ and $p\mapsto\sinh p$ is convex for $p\geq0$.

Christian Blatter
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There might be a geometric interpretation that you are looking for, but I still prefer an algebraic approach. So let's suppose $0 < a < b$, and put $b = ta, t > 1$. Thus: $LHS = \dfrac{1}{\sqrt{ab}} = \dfrac{1}{\sqrt{ta^2}} = \dfrac{1}{a\sqrt{t}}$,and $RHS = \dfrac{\log(at) - \log a}{at- a}= \dfrac{\log t}{a(t-1)}$. Thus you prove: $\dfrac{\log t}{t-1} < \sqrt{t}\iff f(t) =\log t - t\sqrt{t} + \sqrt{t} < 0$. Taking first derivative: $f'(t) = \dfrac{1}{t} - \dfrac{3\sqrt{t}}{2} + \dfrac{1}{2\sqrt{t}} < 0, t > 1\Rightarrow f(t) < f(1) = 0$, and the inequality follows.

DeepSea
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  • How do you pass from $\dfrac{1}{\sqrt{t}} >\dfrac{\log t}{(t-1)}$ to $\dfrac{\log t}{t-1} < \sqrt{t}$ ? – Jean Marie Apr 17 '16 at 21:58
  • I noticed first the problem in your answer, and, in a second step, understood that it came from the question. In such a case, as you are older in mathexchange than me, do we wait that the OP does the work ? – Jean Marie Apr 17 '16 at 22:06
  • It has been done by the OP, I suppose. – Jean Marie Apr 17 '16 at 22:14
  • Excuse me but you should now correct your answer $\dfrac{\log t}{t-1} < \sqrt{t}$ should become $\sqrt{t} \leq \dfrac{t-1}{\log t}$ etc. – Jean Marie Apr 18 '16 at 11:50
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This not an answer, but this remark cannot take place in a simple comment.

It should be said that, under the form $\dfrac{a-b}{ln(a)-ln(b)}$, the logarithmic mean may look rather artificial and difficult to extend to more than 2 variables unless one knows the formula

$$L(a,b)=\int_0^1 \ \varphi(x)dx \ \ \text{with} \ \ \varphi(x)=a^xb^{1-x}=be^{x \ ln(a/b)}$$

which extends to 3D under the form:

$$L(a,b,c)=2\int_{x=0}^1\int_{y=0}^{1-x} a^x \ b^y \ c^{1-x-y} dx dy$$

and more generally in nD in the same way on the simplex defined by $x_1+x_2+\cdots +x_n=1$ (factor $2=2!$ will become $(n-1)!$).

So, the next question is how one can prove for example that

$$G(a,b,c)\leq L(a,b,c) \ ?$$

Jean Marie
  • 88,997