Prove inequality with logarithms:$\log_{a}{\frac{a+b}{2}} + \log_{b}{\frac{a+b}{2}} \ge 2$

Question

Let $ a, b \in (1, \infty)$. Prove that:

$$\log_{a}{(\frac{a+b}{2})} + \log_{b}{(\frac{a+b}{2})} \ge 2$$

I tried switching the bases on each of the logaritm but I got stuck:

$$\frac{\log_{\frac{a+b}{2}}{(ab)}}{\log_{\frac{a+b}{2}}{(a)}\log_{\frac{a+b}{2}}{(b)}}$$

Answer 1

We have to prove that $$\frac{\ln\left(\frac{a+b}{2}\right)}{\ln(a)}+\frac{\ln\left(\frac{a+b}{2}\right)}{\ln(b)}\geq 2$$ this is $$\ln\left(\frac{a+b}{2}\right)(\ln(a)+\ln(b))\geq 2\ln(a)\ln(b)$$ or $$\ln\left(\frac{a+b}{2}\right)\ln(ab)\geq 2\ln(a)\ln(b)$$ Now we have by AM-GM: $$\frac{a+b}{2}\geq \sqrt{ab}$$ taking the logarithm on both sides $$\ln\left(\frac{a+b}{2}\right)\geq \frac{1}{2}\ln(ab)$$ now is $$\ln\left(\frac{a+b}{2}\right)\ln(ab)\geq \frac{1}{2}(\ln(ab))^2$$

Now we have to show that $$(\ln(ab))^2\geq 4\ln(a)\ln(b)$$ this is true since $$(\ln(a)+\ln(b))^2\geq 4\ln(a)\ln(b)$$ this is equivalent to $$(\ln(a)-\ln(b))^2\geq 0$$

Answer 2

Using the change of base formula the inequality is equivalent to

$$ \frac{1}{\ln a}+\frac{1}{\ln b}\ge \frac{2}{\ln\big(\frac{a+b}{2}\big)}. $$

This, in turn, is equivalent to the function

$$ f(x)=\frac{1}{\ln (c-x)}+\frac{1}{\ln x} $$

being minimized on $(1,c-1)$ at $x=c/2$. As $f$ is unbounded (towards $+\infty$) on this interval, if it has only one critical point that will be a minimum. We look to solve

$$ f'(x)= \frac{1}{(c-x)\ln^2(c-x)}-\frac{1}{x\ln^2 x}=0. $$

But $g(x)=x\ln^2 x$ is an increasing function, as it is a product of functions increasing on $(1,\infty$), so the equality $g(c-x)=g(x)$ occurs if and only if $c-x=x$, or $x=c/2$.

Answer 3

Let's set $f(x)=\dfrac 1{\ln(x)}$

We have $f''(x)=\dfrac{2+\ln(x)}{x^2\ln(x)^3}>0$ on $(1,+\infty)$, thus $f$ is convex.

The result is just an application of the convexity inequality

$$f\left(\frac{a+b}2\right)\le \frac 12f(a)+\frac 12f(b)\iff \frac 1{\ln(a)}+\frac 1{\ln(b)}\ge \frac 2{\ln(\frac{a+b}2)}$$

Answer 4

By AM-GM we have $$\frac{a+b}2\geq \sqrt{ab} $$ and since $\log$ is an increasing function for bases $>1$ we have $$\log_a(\frac{a+b}2) +\log_b(\frac{a+b}2) \geq \log_a(\sqrt{ab}) +\log_b(\sqrt{ab}) =1/2(2+\log_a(b)+\log_b(a))$$ And we have that $\log_a(b) +\log_b(a) =\log_a(b) +\frac1{\log_a(b)} \geq 2$ because $x+\frac1 x \geq 2$

Answer 5

By C-S and AM-GM $$\log_a\frac{a+b}{2}+\log_b\frac{a+b}{2}=\frac{1}{\ln_{\frac{a+b}{2}}a}+\frac{1}{\ln_{\frac{a+b}{2}}b}\geq$$ $$\geq\frac{(1+1)^2}{\log_{\frac{a+b}{2}}a+\log_{\frac{a+b}{2}}b}=\frac{4}{\log_{\frac{a+b}{2}}ab}\geq\frac{4}{\log_{\frac{a+b}{2}}\left(\frac{a+b}{2}\right)^2}=2.$$

Answer 6

Using AM >= GM

(a+b)/2 >=sqrt (ab) log(base a) of (( a+b)/2) >= 1/2(1 + log(base a) ( of b)) log(base b) of ((a+b)/2) >= 1/2(1+ log (base b ) (of a )) Adding both LHS >= 1 +(1/2) ( log (base a) (of b) + log ( base b )( of a))

Again using AM>> GM on log terms on RHS we will get

LHS >= 2