MHB How can you prove abc <= 1/8 knowing 1/a + 1/b + 1/c >= 6?

[moonoem]

How can you prove abc <= 1/8 knowing 1/a + 1/b + 1/c >= 6?

 

[kaliprasad]

There are 3 cases
1) if anyone of them are -ve then abc is -ve and so it is true
2) if 2 of them are -ve say a and b then we have $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} >= 6$
$=> - \frac{1}{a} - \frac{1}{b} + \frac{1}{c} >= 6$

so if we can prove that for positive a,b,c if $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} >= 6$ => $abc <=\frac{1}{8}$ we are through

using A.M GM inequality among $\frac{1}{a},\frac{1}{b}, \frac{1}{c} $ we get

$\frac{1}{3} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) >= \sqrt[3]{\frac{1}{abc}}$

or $ 2 >= \sqrt[3]{\frac{1}{abc}}$

or $abc <= \frac{1}{8}$

 

[moonoem]

There are 3 cases
1) if anyone of them are -ve then abc is -ve and so it is true
2) if 2 of them are -ve say a and b then we have $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} >= 6$
$=> - \frac{1}{a} - \frac{1}{b} + \frac{1}{c} >= 6$

so if we can prove that for positive a,b,c if $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} >= 6$ => $abc <=\frac{1}{8}$ we are through

using A.M GM inequality among $\frac{1}{a},\frac{1}{b}, \frac{1}{c} $ we get

$\frac{1}{3} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) >= \sqrt[3]{\frac{1}{abc}}$

or $ 2 >= \sqrt[3]{\frac{1}{abc}}$

or $abc <= \frac{1}{8}$

Honestly, I think there's something with this line:
$ 2 >= \sqrt[3]{\frac{1}{abc}}$
This >= 6
1/3 this >= that (this >= 3that)
It doesn't mean 1/3 × 6 >= that.
this >= 6 and this >= 3that doesn't mean 6 >= 3that or 6 >= 3that.
I forgot to mention a,b,c are positive numbers too
Thanks a ton for the support but something is messed up there.

 

[kaliprasad]

sorry for the mess up

Let me do it again clearly. Now I am unable to reach a solution

 

[MarkFL]

I would use cyclic symmetry (I know this is beyond the scope of the forum in which this problem was posted, but it may prove interesting/informative nonetheless) and say that and critical points (where extrema can occur) will do so for:

$$a=b=c$$

Thus, we may simplify the problem to the objective function (the function we are optimizing)

$$f(a)=a^3$$

subject to the constraint:

$$a\le\frac{1}{2}$$

And so computing the derivative of the objective function (to learn how it behaves) we find:

$$f'(a)=3a^2$$

We know that for all \(a\) satisfying the constraint, we have:

$$f'(a)>0$$

This means over the relevant domain given by the constraint, our objective function is strictly increasing, and so we may conclude that the maximum value of the function is:

$$f_{\max}=f\left(\frac{1}{2}\right)=\frac{1}{8}$$

And so:

$$f(a)\le\frac{1}{8}$$

This is what we were asked to show. I haven't looked at using the AM-GM inequality (which I know is also good for problems like this), but I'm confident Kali will make it work.

 

[moonoem]

I would use cyclic symmetry (I know this is beyond the scope of the forum in which this problem was posted, but it may prove interesting/informative nonetheless) and say that and critical points (where extrema can occur) will do so for:

$$a=b=c$$

Thus, we may simplify the problem to the objective function (the function we are optimizing)

$$f(a)=a^3$$

subject to the constraint:

$$a\le\frac{1}{2}$$

And so computing the derivative of the objective function (to learn how it behaves) we find:

$$f'(a)=3a^2$$

We know that for all \(a\) satisfying the constraint, we have:

$$f'(a)>0$$

This means over the relevant domain given by the constraint, our objective function is strictly increasing, and so we may conclude that the maximum value of the function is:

$$f_{\max}=f\left(\frac{1}{2}\right)=\frac{1}{8}$$

And so:

$$f(a)\le\frac{1}{8}$$

This is what we were asked to show. I haven't looked at using the AM-GM inequality (which I know is also good for problems like this), but I'm confident Kali will make it work.

Thanks for caring and for publishing this interesting post... But it seems to be indeed out of my knowledge and the range of knowledge I am allowed to use.
I would really appreciate if someone could solve it with a method that suits me better.

 

[moonoem]

Given a,b,c > 0 satisfying 1/(1+a) + 1/(1+b) + 1/(1+c) >= 2. Prove that abc <= 1/8This is the full problem. This one is just my attempt for this full problem, and it is incorrect. Sorry for making some of you lose some braincells unnecessarily.

 

[kaliprasad]

Assuming that a,b,c are positive
If we have $\frac{1}{1+a } + \frac{1}{1+b} + \frac{1}{1+c} = 2$
Hence
$(1+b)(1+c) + (1+c)(1+a) + (1+a)(1+b) = 2(1+a)(1+b)(1+c)$
or $1 + b + c + bc + c + a + ac + 1+ a + b + ab = 2 + 2a + 2b + 2c + 2ab + 2bc + 2ac + 2abc$
or $2abc + ab + bc + ca = 1$
Applying AM GM inequality we get
$\frac{1}{4}(ab+bc+ca+2abc) >= \sqrt[4]{2abc.ab.bc.ca}$
Or $\frac{1}{4} >= \sqrt[4] {2 a^3b^3c^3}$
Or $\frac{1}{256} >= 2(abc)^3$
or $(abc) <= \sqrt[3]\frac{1}{512}$
or $(abc) <= \frac{1}{8}$

Now for the If we have $\frac{1}{1+a } + \frac{1}{1+b} + \frac{1}{1+c} >= 2$

one or more of denominator has to be smaller and hence one of a,b or c shall be smaller and hence the product shall be even lesser

 

[moonoem]

Assuming that a,b,c are positive
If we have $\frac{1}{1+a } + \frac{1}{1+b} + \frac{1}{1+c} = 2$
Hence
$(1+b)(1+c) + (1+c)(1+a) + (1+a)(1+b) = 2(1+a)(1+b)(1+c)$
or $1 + b + c + bc + c + a + ac + 1+ a + b + ab = 2 + 2a + 2b + 2c + 2ab + 2bc + 2ac + 2abc$
or $2abc + ab + bc + ca = 1$
Applying AM GM inequality we get
$\frac{1}{4}(ab+bc+ca+2abc) >= \sqrt[4]{2abc.ab.bc.ca}$
Or $\frac{1}{4} >= \sqrt[4] {2 a^3b^3c^3}$
Or $\frac{1}{256} >= 2(abc)^3$
or $(abc) <= \sqrt[3]\frac{1}{512}$
or $(abc) <= \frac{1}{8}$

Now for the If we have $\frac{1}{1+a } + \frac{1}{1+b} + \frac{1}{1+c} >= 2$

one or more of denominator has to be smaller and hence one of a,b or c shall be smaller and hence the product shall be even lesser

Honestly, this is an entire new method to solve for me. And I love it! I will ask my teacher if this method is usable. Thanks a lot for everyone's help all those time!