Can AM-GM Inequality Solve This Algebraic Problem?

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In summary: In this case, we can see that the equality does not hold, so the inequality is not true. In summary, your use of the AM-GM inequality is correct, but your solution is not correct.
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Homework Statement



{(x+y+z)^3-2(x+y+z)(x^2+y^2+z^2)}/xyz ≤ 9

Homework Equations



AM-GM inequality x+y+z ≥ 3(√xyz)(cube root) and xy+yz+zx ≥ 3√(xyz)^2(cube root)

The Attempt at a Solution



This is my attempt but I don't know if I am using the AM-GM inequality correctly

(x+y+z)^3/xyz ≤ 9+ [2(x+y+z)(x^2+y^2+z^2)]/xyz

(x+y+z)^3/xyz ≤ [9xyz+ 2(x+y+z){(x+y+z)(x+y+z)-2(xy+yz+zx)}]/xyz

(3√xyz)3/xyz ≤ [9xyz +2(x+y+z)^3 -4(x+y+z)(xy+yz+zx)]/xyz
27xyz/xyz ≤ [9xyz +2(3√xyz)^3 -4(3√xyz)(3√(xyz)^2)]/xyz

27xyz/xyz ≤ [9xyz +54xyz -36xyz]/xyz
27 ≤ 27

Is this correct?
 
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  • #2


I would like to provide some feedback on your solution. The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. In other words, for any set of positive numbers, the sum of the numbers divided by the number of numbers is greater than or equal to the product of the numbers raised to the power of 1/number of numbers. In this case, we have three numbers (x, y, and z), so the inequality would be:

(x+y+z)/3 ≥ (xyz)^(1/3)

Using this inequality, we can rewrite the original expression as:

(x+y+z)^3/xyz ≤ 9 + 2(x+y+z)(x^2+y^2+z^2)/xyz

Applying the AM-GM inequality to the first term on the right side, we get:

(x+y+z)^3/xyz ≤ 9 + 2(x+y+z)(x^2+y^2+z^2)/xyz ≤ 9 + 2(3√xyz)(xy+yz+zx)/xyz

Using the AM-GM inequality again on the second term on the right side, we get:

(x+y+z)^3/xyz ≤ 9 + 2(3√xyz)(xy+yz+zx)/xyz ≤ 9 + 2(3√(xyz)^2)(x+y+z)/xyz

Simplifying, we get:

(x+y+z)^3/xyz ≤ 9 + 6(x+y+z)/xyz

Now, we can use the fact that (x+y+z)/3 ≥ (xyz)^(1/3) to rewrite the right side as:

(x+y+z)^3/xyz ≤ 9 + 2(xyz)^(1/3)

Finally, we can use the fact that (xyz)^(1/3) = (xyz)^2/3 to get:

(x+y+z)^3/xyz ≤ 9 + 2(xyz)^2/3

This is as far as we can simplify the expression using the AM-GM inequality. We can see that the expression on the right side is not equal to 9, so the original inequality is not true. This means that your attempt is not correct. However, your use of the AM-GM inequality is correct. Keep in mind that the AM-GM inequality does not always guarantee that the equality
 

What is the AM-GM inequality problem?

The AM-GM inequality problem is a mathematical concept that states that the Arithmetic Mean (AM) of a set of numbers is always greater than or equal to the Geometric Mean (GM) of the same set of numbers. This means that the average of a set of numbers will always be equal to or greater than the product of those numbers.

Why is the AM-GM inequality problem important?

The AM-GM inequality problem is important because it has many applications in mathematics and other fields such as economics, physics, and engineering. It is used to prove other mathematical theorems and inequalities, and it has practical applications in finding the optimal solution to problems involving maximizing or minimizing a set of values.

How is the AM-GM inequality problem solved?

The AM-GM inequality problem can be solved using various mathematical techniques, such as algebraic manipulation, calculus, and proof by induction. The specific method used will depend on the context and complexity of the problem being solved.

What are some real-life examples of the AM-GM inequality problem?

The AM-GM inequality problem can be applied to various real-life scenarios, such as finding the most efficient distribution of resources in a business, maximizing profits in a market, or determining the minimum amount of material needed to construct a structure. It can also be used to solve problems in physics, such as finding the minimum energy needed to perform a task.

Is there a generalization of the AM-GM inequality problem?

Yes, there is a generalization of the AM-GM inequality problem known as the Power Mean inequality. It states that for any positive real numbers, the p-th Power Mean is greater than or equal to the q-th Power Mean if p > q. This generalization allows for a wider range of applications and can be used to prove other mathematical inequalities.

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