Hi! Today i was trying to prove an interesting inequality, but no matter what approach i was choosing, i couldnt succed.
The inequality is logaa+b2ab×logba+b2ab≥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.
Hi! Today i was trying to prove an interesting inequality, but no matter what approach i was chosing, i couldnt succed.
The inequality is logaa+b2ab×logba+b2ab≥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
0<y<z<1⟹∃x>1 such that zx=y⟹logz(y)=xlogz(z)>x∗1>1.
In short, 0<y<z<1⟹logz(y)>1, which is weird.
Similarly, 0<z<y<1⟹1>logz(y)>0, which is weird but consistent.
And finally, 0<z<1<y⟹logz(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.
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 : ab≥a+b2ab (1).
Referring to logarithms, we know that for a>0, if x≥y then logax≥logby, but if 0<a<1 and x≥y, then logax≤logby (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:
logaab≤logaa+b2ab and logbab≤logba+b2ab
Multiplying these 2 inequalities, we have:
logaa+b2ab×logba+b2ab≥logaab×logaab
So the problem is now to prove that logaab×logaab≥1
Let's simplify this product using some log properties:
Now let logab=x, then by log properties,logba=x1
We can prove that x>0 (so 1/x will be also >0), because x=logab=lgblga, 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.
So now we need to prove that 41(2+x+x1)≥1 and we are done.
For this, let's use a particular version of AM-GM inequality: ba+ab≥2, (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: 1x+x1≥2⇔x+x1≥2⇔2+x+x1≥4⇔41(2+x+x1)≥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.
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 : ab≥a+b2ab (1).
Referring to logarithms, we know that for a>0, if x≥y then logax≥logby, but if 0<a<1 and x≥y, then logax≤logby (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:
logaab≤logaa+b2ab and logbab≤logba+b2ab
Multiplying these 2 inequalities, we have:
logaa+b2ab×logba+b2ab≥logaab×logaab
So the problem is now to prove that logaab×logaab≥1
Let's simplify this product using some log properties:
Now let logab=x, then by log properties,logba=x1
We can prove that x>0 (so 1/x will be also >0), because x=logab=lgblga, 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.
So now we need to prove that 41(2+x+x1)≥1 and we are done.
For this, let's use a particular version of AM-GM inequality: ba+ab≥2, (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: 1x+x1≥2⇔x+x1≥2⇔2+x+x1≥4⇔41(2+x+x1)≥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.
I have reposted soyour post is more legible. I am still studying it.
ab≥a+b2ab (1).
Referring to logarithms, we know that for a>0, if x≥y then logax≥logby, but if 0<a<1 and x≥y, then logax≤logby (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:
logaab≤logaa+b2ab and logbab≤logba+b2ab
Multiplying these 2 inequalities, we have:
logaa+b2ab×logba+b2ab≥logaab×logaab
So the problem is now to prove that logaab×logaab≥1
Let's simplify this product using some log properties:
Now let logab=x, then by log properties,logba=x1
We can prove that x>0 (so 1/x will be also >0), because x=logab=lgblga, 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.
So now we need to prove that 41(2+x+x1)≥1 and we are done.
For this, let's use a particular version of AM-GM inequality: ba+ab≥2, (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: 1x+x1≥2⇔x+x1≥2⇔2+x+x1≥4⇔41(2+x+x1)≥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.
I have reposted soyour post is more legible. I am still studying it.
ab≥a+b2ab (1).
Referring to logarithms, we know that for a>0, if x≥y then logax≥logby, but if 0<a<1 and x≥y, then logax≤logby (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:
logaab≤logaa+b2ab and logbab≤logba+b2ab
Multiplying these 2 inequalities, we have:
logaa+b2ab×logba+b2ab≥logaab×logaab
So the problem is now to prove that logaab×logaab≥1
Let's simplify this product using some log properties:
Now let logab=x, then by log properties,logba=x1
We can prove that x>0 (so 1/x will be also >0), because x=logab=lgblga, 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.
So now we need to prove that 41(2+x+x1)≥1 and we are done.
For this, let's use a particular version of AM-GM inequality: ba+ab≥2, (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: 1x+x1≥2⇔x+x1≥2⇔2+x+x1≥4⇔41(2+x+x1)≥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.
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