# Inequality Proof: Showing \left|ab\right|\leq\frac{1}{2}(a^{2}+b^{2})

• kpoltorak
In summary: The two cases that you mention both hold true. Thank you for clearing that up for me.In summary, the homework statement is thattriangle inequality seems to be useless. The Attempt at a Solution states that (a+b)^{2}=a^{2}+b^{2}+2ab, which is true if both terms in the parentheses are positive.
kpoltorak

## Homework Statement

Show that $$\forall a,b \in R$$:
$$\left|ab\right|\leq\frac{1}{2}(a^{2}+b^{2})$$

## Homework Equations

Triangle Inequality seems to be useless.

## The Attempt at a Solution

$$(a+b)^{2}=a^{2}+b^{2}+2ab$$
$$2ab=(a+b)^{2}-(a^{2}+b^{2})$$
$$ab=\frac{1}{2}(a+b)^{2}-\frac{1}{2}(a^{2}+b^{2})$$
$$\left|ab\right|=\left|\frac{1}{2}(a+b)^{2}-\frac{1}{2}(a^{2}+b^{2})\right|$$
$$\left|ab\right|=\left|\frac{1}{2}(a^{2}+b^{2})-\frac{1}{2}(a+b)^{2}\right|$$

Well you pretty much have it. Take a look at the right hand side, they all consist of squares, so...?

Right I've noticed that the values on both sides of the minus sign are all positive, however that doesn't necessarily mean that $$\left|ab\right|$$ is less than $$\frac{1}{2}(a^{2}+b^{2})$$. Because its an absolute value, the LHS of the minus sign could be smaller than the RHS while preserving the equality. For example, it is NOT TRUE that $$\left|6\right|=\left|2-8\right|\rightarrow 6\leq 2$$

Well yes I realize that, but you seem to have forgotten what the question is. Take a look at it again!
$$\left|ab\right|\leq\frac{1}{2}(a^{2}+b^{2})$$

Notice how the right side only has that one term in there, and by the way, $$\frac{1}{2}(a+b)^2\geq 0$$

I think I see...?

$$|ab|\leq\left|\frac{1}{2}(a^{2}+b^{2})-\frac{1}{2}(a+b)^{2}+\frac{1}{2}(a+b)^{2}\right|$$
$$\left|ab\right|\leq\left|\frac{1}{2}(a^{2}+b^{2})\right|$$
$$\left|ab\right|\leq\frac{1}{2}(a^{2}+b^{2})$$

You shouldn't have the term you added inside the absolute value sign. For example,

$$|10|=|5-15|$$

$$|10|\leq |5-15+15|=|5|$$ is obviously wrong.

Instead, you should consider both cases when the RHS inside the absolute value is more than zero, and then less than zero and show both cases hold true for the inequality you want to prove.

Now I see. Thank you for all your help!

You're welcome

Here was the way I thought about it:
$$|ab| \leq \frac{1}{2}(a^2+b^2)$$
times by 2 and square both sides

$$4a^2b^2 \leq (a^2+b^2)^2$$

$$0 \leq a^4 - 2a^2b^2 + b^4$$

$$0 \leq (a^2-b^2)^2$$

and since it is squared it must be greater than or equal to zero

Last edited:
That's much more elegant Tom.

## 1. What is the purpose of the inequality proof, \left|ab\right|\leq\frac{1}{2}(a^{2}+b^{2})?

The purpose of this inequality proof is to demonstrate that the product of two real numbers, |ab|, is always less than or equal to half the sum of their squares, \frac{1}{2}(a^{2}+b^{2}). This proof is important in mathematics and physics as it helps to establish the relationship between multiplication and addition of real numbers.

## 2. How is this inequality proof derived?

This inequality proof can be derived using the properties of real numbers, such as the distributive, associative, and commutative properties. It also involves the use of the Pythagorean theorem and the concept of absolute value. The proof can be further expanded and generalized using algebraic manipulations and mathematical induction.

## 3. What are the key assumptions made in this inequality proof?

The key assumptions made in this inequality proof are that a and b are real numbers, and that they can be positive, negative, or zero. The proof also assumes the basic properties of real numbers, such as the existence of inverses and the closure under addition and multiplication.

## 4. Can this inequality proof be applied to complex numbers?

No, this inequality proof only applies to real numbers. Complex numbers have different properties and do not follow the same rules as real numbers. In fact, the inequality \left|ab\right|\leq\frac{1}{2}(a^{2}+b^{2}) does not hold for all complex numbers a and b.

## 5. How is this inequality proof relevant in practical applications?

This inequality proof has various practical applications in fields such as economics, physics, and statistics. It helps in understanding and solving problems involving multiplication and addition of real numbers, and has applications in inequality analysis, optimization, and proof of inequalities in other mathematical theorems.

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