I think i got it!
Back to my idea with means, we know that geometric mean is greater or equal to harmonic mean, so for 2 numbers a and b, we will have : \(\displaystyle \displaystyle\sqrt{ab}\geq \frac{2ab}{a+b} \text{ (1)}\).
Referring to logarithms, we know that for \(\displaystyle a>0, \text{ if } x\geq y \text{ then } \log_a{x}\geq\log_b{y}\), but if \(\displaystyle 0<a<1 \text{ and } x\geq y, \text{ then } \log_a{x}\leq\log_b{y} \text{ (2)}\) .. 2
Based on (2), we can write 2 inequalities from (1), by taking log_a for the first inequality and log_b for the second. As a and b takes values from 0 to 1, by taking the logarithm base a and b, we will change greater or equal to less or equal. So we get:
\(\displaystyle \displaystyle \log_a{\sqrt{ab}} \leq \log_a{\frac{2ab}{a+b}} \text { and } \log_b{\sqrt{ab}} \leq \log_b{\frac{2ab}{a+b}} \)
Multiplying these 2 inequalities, we have:
\(\displaystyle \displaystyle \log_a{\frac{2ab}{a+b}} \times \log_b{\frac{2ab}{a+b}}\geq \log_a{\sqrt{ab}}\times \log_a{\sqrt{ab}}\)
So the problem is now to prove that \(\displaystyle \log_a{\sqrt{ab}}\times \log_a{\sqrt{ab}}\geq 1\)
Let's simplify this product using some log properties:
\(\displaystyle \displaystyle\log_a{\sqrt{ab}}\times \log_a{\sqrt{ab}} = \frac 12\log_a{(ab)}\times \frac 12\log_a{(ab)}=\frac 14\log_a{(ab)}\times\log_a{(ab)}=\frac 14(\log_a{a}+\log_a{b})(\log_b{b}+\log_b{a})=\frac 14(1+\log_a{b})(1+\log_b{a})\)
Now let \(\displaystyle \displaystyle\log_a{b}=x, \text{ then by log properties}, \log_b{a}=\frac 1x\)
We can prove that x>0 (so 1/x will be also >0), because \(\displaystyle \displaystyle x=\log_a{b}=\frac{\lg{a}}{\lg{b}}\), as we have that 0<a<1 and 0<b<1, then lg(a)<0 and lg(b)<0, so their division, will give us the + sign.
Introducing x, we get:
\(\displaystyle \displaystyle\log_a{\sqrt{ab}}\times \log_a{\sqrt{ab}} = \frac 14(1+x)(1+\frac 1x)=\frac 14(1+x+\frac 1x+1)=\frac 14(2+x+\frac 1x)\)
So now we need to prove that \(\displaystyle \displaystyle \frac14(2+x+\frac 1x)\geq1\) and we are done.
For this, let's use a particular version of AM-GM inequality: \(\displaystyle \displaystyle\frac ab+\frac ba\geq 2, \text{ (that is true for every a,b>0)}\)
For our case, let a=x and b=1. I proved that x>0 so we can use this inequality: \(\displaystyle \displaystyle \frac x1+\frac 1x\geq 2\Leftrightarrow x+\frac 1x\geq 2 \Leftrightarrow 2+x+\frac 1x \geq 4 \Leftrightarrow \frac 14(2+x+\frac 1x)\geq 1\)
So it's done!
Now i have some questions about my prove, as i can't explain some things so it's possible that this prove is bad.
1)Am i allowed to multiply the ineqaulities in the first part of the proof? as logs can be negative, and if it's the case i was wondering if it's change the sign.
2)Can someone give an indication on how to prove the (2) inequality? as i think it's obvious, but i'd like to see how it can be proved.