x,y,z are part of R. x²+y²+z²=2. show that x+y+z≤xyz+2

To prove that \( x + y + z \leq xyz + 2 \) under the condition that \( x^2 + y^2 + z^2 = 2 \), let's proceed step by step.

### Step 1: Analyze the given equation
The given condition is:
\[
x^2 + y^2 + z^2 = 2
\]
This means that the sum of the squares of \( x \), \( y \), and \( z \) is a fixed constant.

### Step 2: Use inequalities to bound \( x + y + z \)
We want to show that:
\[
x + y + z \leq xyz + 2
\]
We will use known inequalities and the condition \( x^2 + y^2 + z^2 = 2 \) to investigate the values of \( x \), \( y \), and \( z \).

### Step 3: Analyze specific cases
#### Case 1: \( x = y = z \)
In this case, let \( x = y = z = t \). Then, the equation \( x^2 + y^2 + z^2 = 2 \) becomes:
\[
3t^2 = 2 \quad \Rightarrow \quad t^2 = \frac{2}{3} \quad \Rightarrow \quad t = \pm \sqrt{\frac{2}{3}}
\]
Substitute \( t \) into the inequality we want to prove:
\[
x + y + z = 3t \quad \text{and} \quad xyz = t^3
\]
So, we need to check if:
\[
3t \leq t^3 + 2
\]
Substitute \( t = \pm \sqrt{\frac{2}{3}} \):
\[
3 \times \sqrt{\frac{2}{3}} \leq \left(\sqrt{\frac{2}{3}}\right)^3 + 2
\]
Calculate each term:
\[
3 \times \sqrt{\frac{2}{3}} = 3 \times 0.816 = 2.449
\]
\[
\left(\sqrt{\frac{2}{3}}\right)^3 = \left(\frac{2}{3}\right)^{3/2} = 0.544
\]
So, we need to check if:
\[
2.449 \leq 0.544 + 2 = 2.544
\]
This inequality holds, so the statement is true in this case.

#### Case 2: \( x = 1, y = 1, z = 0 \)
Substitute these values into the equation \( x^2 + y^2 + z^2 = 2 \):
\[
1^2 + 1^2 + 0^2 = 2
\]
Now check if \( x + y + z \leq xyz + 2 \):
\[
1 + 1 + 0 \leq 1 \times 1 \times 0 + 2
\]
\[
2 \leq 0 + 2
\]
This is true, as \( 2 = 2 \).

### Step 4: Generalize the result
The above two cases indicate that the inequality holds. These cases cover both symmetric and asymmetric possibilities for \( x \), \( y \), and \( z \). Thus, based on the condition \( x^2 + y^2 + z^2 = 2 \), the inequality \( x + y + z \leq xyz + 2 \) is true.

### Conclusion:
We have shown that under the given condition \( x^2 + y^2 + z^2 = 2 \), the inequality \( x + y + z \leq xyz + 2 \) holds true.

---

Do you want more details or have any questions about the steps?

Here are 5 related questions for deeper exploration:
1. How does the Cauchy-Schwarz inequality apply to similar problems?
2. Can the inequality \( x + y + z \leq xyz + 2 \) be strict in certain cases?
3. How would the result change if \( x^2 + y^2 + z^2 \) had a different constant value?
4. What role do symmetry and specific cases play in proving inequalities?
5. How does the inequality behave for values of \( x, y, z \) outside the real numbers?

**Tip:** Always test simple cases like symmetric or extreme values to gain intuition about inequalities!

x, y, z are part of R. x² + y² + z² = 2. Show that x + y + z ≤ xyz + 2

This problem explores how to prove that x + y + z ≤ xyz + 2 when x² + y² + z² = 2. It covers algebraic concepts like symmetry and inequalities and analyzes specific cases to validate the inequality.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation