Logarithmic inequality: log_a[(2ab)/(a+b)] * log_b[(2ab)/(a+b)] >= 1

[Stas]

Hi! Today i was trying to prove an interesting inequality, but no matter what approach i was choosing, i couldnt succed.

The inequality is \(\displaystyle \displaystyle \log_a {\frac{2ab}{a+b}}\times\log_b {\frac{2ab}{a+b}}\geq {1}\) and i should prove that it's true for a,b from 0 to 1.

First i tried to simplify these logarithms, but this didn't help. Then i realised that 2ab/(a+b) is the harmonic mean for n=2 and i used some inequalities based on means, but this also didn't help.

 

[JeffM]

Hi! Today i was trying to prove an interesting inequality, but no matter what approach i was chosing, i couldnt succed.

The inequality is \(\displaystyle \displaystyle \log_a {\frac{2ab}{a+b}}\times\log_b {\frac{2ab}{a+b}}\geq {1}\) and i should prove that it's true for a,b from 0 to 1.

First i tried to simplify these logarithms, but this didn't help. Then i realised that 2ab/(a+b) is the harmonic mean for n=2 and i used some inequalities based on means, but this also didn't help.

I have played around with this algebraically. The fundamental problem algebraically is that, although both factors are necessarily positive, if one factor exceeds 1, the other factor is exceeded by 1. Thus, it is not easy to determine if the product exceeds or is exceeded by 1.

Futhermore, dealing with logarithms to a base between zero and one is unusual.

Consider

\(\displaystyle 0 < y < z < 1 \implies \exists \ x > 1 \text { such that } z^x = y \implies log_z(y) = x log_z(z) > x * 1 > 1.\)

In short, \(\displaystyle 0 < y < z < 1 \implies log_z(y) > 1,\) which is weird.

Similarly, \(\displaystyle 0 < z < y < 1 \implies 1 > log_z(y) > 0,\) which is weird but consistent.

And finally, \(\displaystyle 0 < z < 1 < y \implies log_z(y) < 0.\)

Our (or at least my) intuition about logs is scrambled when dealing with fractional bases. And here we may have two such bases.

I shall work some more over the next few days. My next approach will be to convert everything to base e and then apply differential calculus. That may work where mere algebra seems to fail.

 

[Stas]

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 \(\displaystyle 0<a<1 \text{ and } x\geq y, \text{ then } \log_a{x}\leq\log_b{y} \text{ (2)}\)

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.

 

[Deleted member 4993]

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.

Are you referring to what you called the AM-GM inequality? Or the inequality invoked at the top

 

[Stas]

Are you referring to what you called the AM-GM inequality? Or the inequality invoked at the top

At the top, please.

 

[JeffM]

I have reposted so your post is more legible. I am still studying it.

\(\displaystyle \displaystyle\sqrt{ab}\geq \frac{2ab}{a+b} \text{ (1)}\).

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 \(\displaystyle 0<a<1 \text{ and } x\geq y, \text{ then } \log_a{x}\leq\log_b{y} \text{ (2)}\)

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:

\(\displaystyle \displaystyle\log_a{\sqrt{ab}}\times \log_a{\sqrt{ab}} = \frac 12\log_a{(ab)}\times \frac 12\log_a{(ab)}=\frac 14\log_a{(ab)}\times\log_a{(ab)}=\)

\(\displaystyle \frac 14(\log_a{a}+\log_a{b})(\log_b{b}+\log_b{a})=\frac 14(1+\log_a{b})(1+\log_b{a})\)

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. YOU CAN'T DO THIS. b IS ASSUMED < 1 BY HYPOTHESIS. AND LOG TO THE BASE b = 1 IS MEANINGLESS. BUT THIS ERROR MAY BE HARMLESS. NOT SURE.

I proved that x>0 so we can use this inequality: \(\displaystyle \displaystyle \frac x1+\frac 1x\geq 2\Leftrightarrow x+\frac 1x\geq 2 \Leftrightarrow 2+x+\frac 1x \geq 4 \Leftrightarrow \frac 14(2+x+\frac 1x)\geq 1\)


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

Inequality 2 is true, and I can prove it. But we also get that the logarithms of numbers greater than 1 are negative. I am not sure how that will affect your proof.

 

[Stas]

I have reposted so your post is more legible. I am still studying it.

\(\displaystyle \displaystyle\sqrt{ab}\geq \frac{2ab}{a+b} \text{ (1)}\).

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 \(\displaystyle 0<a<1 \text{ and } x\geq y, \text{ then } \log_a{x}\leq\log_b{y} \text{ (2)}\)

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:

\(\displaystyle \displaystyle\log_a{\sqrt{ab}}\times \log_a{\sqrt{ab}} = \frac 12\log_a{(ab)}\times \frac 12\log_a{(ab)}=\frac 14\log_a{(ab)}\times\log_a{(ab)}=\)

\(\displaystyle \frac 14(\log_a{a}+\log_a{b})(\log_b{b}+\log_b{a})=\frac 14(1+\log_a{b})(1+\log_b{a})\)

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. YOU CAN'T DO THIS. b IS ASSUMED < 1 BY HYPOTHESIS. AND LOG TO THE BASE b = 1 IS MEANINGLESS. BUT THIS ERROR MAY BE HARMLESS. NOT SURE.

I proved that x>0 so we can use this inequality: \(\displaystyle \displaystyle \frac x1+\frac 1x\geq 2\Leftrightarrow x+\frac 1x\geq 2 \Leftrightarrow 2+x+\frac 1x \geq 4 \Leftrightarrow \frac 14(2+x+\frac 1x)\geq 1\)


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

Inequality 2 is true, and I can prove it. But we also get that the logarithms of numbers greater than 1 are negative. I am not sure how that will affect your proof.

About the part "For our case, let a=x and b=1", the a and b from inequality a/b+b/a>=2, have nothing to do with a and b from the problem. Sorry, my bad, i should have used other variables as c and d or i could used from the begining the inequality x+1/x>=2, and this will be great too.