- #1

moonoem

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

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In summary, The problem asks to prove that if a,b,c are positive, then abc <= 1/8. Using cyclic symmetry and the AM-GM inequality, it is proved that if a,b,c are positive, then abc <= 1/8.

- #1

moonoem

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

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- #2

kaliprasad

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

- #3

moonoem

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kaliprasad said:

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.

- #4

kaliprasad

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sorry for the mess up

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

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

Last edited:

- #5

MarkFL

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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:

\(\displaystyle a=b=c\)

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

\(\displaystyle f(a)=a^3\)

subject to the constraint:

\(\displaystyle a\le\frac{1}{2}\)

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

\(\displaystyle f'(a)=3a^2\)

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

\(\displaystyle f'(a)>0\)

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:

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

And so:

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

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.

\(\displaystyle a=b=c\)

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

\(\displaystyle f(a)=a^3\)

subject to the constraint:

\(\displaystyle a\le\frac{1}{2}\)

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

\(\displaystyle f'(a)=3a^2\)

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

\(\displaystyle f'(a)>0\)

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:

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

And so:

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

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.

Last edited:

- #6

moonoem

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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.MarkFL said:

\(\displaystyle a=b=c\)

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

\(\displaystyle f(a)=a^3\)

subject to the constraint:

\(\displaystyle a\le\frac{1}{2}\)

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

\(\displaystyle f'(a)=3a^2\)

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

\(\displaystyle f'(a)>0\)

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:

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

And so:

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

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.

I would really appreciate if someone could solve it with a method that suits me better.

- #7

moonoem

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- #8

kaliprasad

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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

- #9

moonoem

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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!kaliprasad said:

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

The proof for this statement involves using the AM-GM inequality, which states that the arithmetic mean of a set of numbers is always greater than or equal to the geometric mean of those numbers. In this case, we can use the AM-GM inequality to show that 1/a + 1/b + 1/c >= 6 is equivalent to abc <= 1/8.

The AM-GM inequality is a fundamental principle in mathematics that states that the arithmetic mean of a set of numbers is always greater than or equal to the geometric mean of those numbers. This means that the sum of a set of numbers divided by the number of terms is always greater than or equal to the nth root of the product of those numbers, where n is the number of terms.

In the given statement, we have 1/a + 1/b + 1/c >= 6, which can be rewritten as (1/a + 1/b + 1/c)/3 >= (1/a * 1/b * 1/c)^(1/3). By the AM-GM inequality, the left side is always greater than or equal to the right side. Therefore, (1/a * 1/b * 1/c)^(1/3) <= 6/3 = 2. This means that the product of 1/a, 1/b, and 1/c must be less than or equal to 2^3 = 8. Since abc is the product of these three numbers, we can conclude that abc <= 1/8.

Yes, the AM-GM inequality is a very versatile inequality and can be used in many different situations. It is commonly used in algebra, calculus, and geometry to prove various mathematical statements and solve problems. It is also used in physics, economics, and other fields of science to model and analyze real-world phenomena.

Yes, there are other ways to prove this statement. One possible approach is to use the Cauchy-Schwarz inequality, which states that the square of the sum of two numbers is always greater than or equal to the sum of the squares of those numbers. By applying this inequality to the given statement, we can show that abc <= 1/8. Another approach is to use the quadratic formula to solve for the roots of the equation 1/a + 1/b + 1/c = 6, and then use the properties of quadratic equations to prove that abc <= 1/8.

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