Introduction to Inequalities Involving Sum of Squares and Products

Mathematics is a rich field where inequalities play a crucial role in understanding the relationships between different mathematical quantities. This article will explore the proofs and generalization of inequalities involving the sum of squares and products. Specifically, we will illustrate how to show that ({a}^2{b}^2 - ab b^2 - b^2 geq 0) and how this can be extended to a more general result.

Proof of (a^2b^2 - ab b^2 - b^2 geq 0)

To prove the inequality (a^2b^2 - ab b^2 - b^2 geq 0), we can start by simplifying it:

[begin{align*}displaystyle a^2b^2 - ab b^2 - b^2 a^2b^2 - ab a^2b^2 - 2ab b^2 ab - b^2 (ab - b)^2 ab - b^2end{align*}]

This expression can be reorganized as follows:

[begin{align*}(ab - b)^2 ab - b^2 (ab - b)(ab - b) b(a - b) ab(a - b) - b^2(a - b) b(a - b) a^2b - ab^2 - b^3 ab^2 ab - b^2 a^2b - b^2 - b^2 ab - ab b^2 a^2b - b^2end{align*}]

Given that we have an expression in the form of ((ab - b)^2 (ab - b^2)), which is always nonnegative because it is the sum of two squares, we conclude that:

[a^2b^2 - ab b^2 - b^2 geq 0]

Generalization to (sum a^2 geq sum ab)

Now, let's consider the more general inequality (a^2 b^2 c^2 geq ab bc ac). We can use a substitution and the concept of a quadratic function to prove this result.

Using Substitution and Quadratic Functions

Suppose we want to prove (a^2 b^2 c^2 geq ab bc ac) under the condition (c ab). In this case, we can substitute (c ab) and proceed with the proof.

First, consider the inequality as a quadratic in (a):

[a^2 - (b c)a b^2 c^2 - bc geq 0]

This can be rewritten as:

[a^2 - (b ab)a b^2 (ab)^2 - b(ab) geq 0]

Simplifying, we get:

[a^2 - ab - a^2b b^2 a^2b^2 - ab^2 geq 0]

This expression can be further simplified and analyzed using the discriminant of the quadratic equation to determine whether it is always nonnegative.

Discriminant Analysis of the Quadratic Function

The discriminant (Delta) of the quadratic equation (s a^2 t a u 0) is given by:

[Delta t^2 - 4su]

In our case, the quadratic equation is:

[a^2 - (b ab)a b^2 (ab)^2 - b(ab) 0]

The coefficients are:

[s 1, quad t -(b ab), quad u b^2 a^2b^2 - ab^2]

Then the discriminant becomes:

[Delta (b ab)^2 - 4(b^2 a^2b^2 - ab^2)]

Expanding and simplifying the discriminant:

[Delta b^2 2ab^2 a^2b^2 - 4b^2 - 4a^2b^2 4ab^2]

[Delta 6ab^2 - 3a^2b^2 - 3b^2]

This expression can be further simplified to:

[Delta -3b^2(a^2 b^2 - 2ab 1)]

[Delta -3b^2(a - b)^2]

Since ((a - b)^2 geq 0), we have that (Delta leq 0). Therefore, the quadratic equation is either always nonpositive or always nonnegative. Given that ((a, b, c ab)) are real numbers, we conclude that the quadratic is always nonnegative.

General Result: (sum a^2 geq sum ab)

By the proof above, we can generalize that for any real numbers (a, b, c), the inequality:

[a^2 b^2 c^2 geq ab bc ac]

holds, except when (a b c 0), in which case equality holds.

Conclusion: We have shown that both the initial inequality and the more general result hold under specific conditions.

Conclusion

By using algebraic manipulation, substitution, and the properties of quadratic functions, we have proven the given inequalities. The proof demonstrates the power of algebraic techniques in handling such problems and highlights the importance of understanding the discriminant in quadratic functions.