https://file.aiquickdraw.com/mathbot/file/1730296069014ufbvuqjd.jpg

The inequality in the image is:

\[
x^2 + \psi^2 \geq \frac{1}{2} (x + \psi)^2
\]

To analyze this, let's expand and simplify the inequality.

### Step 1: Expand the Right Side
Expanding \((x + \psi)^2\):
\[
(x + \psi)^2 = x^2 + 2x\psi + \psi^2
\]

Thus, the inequality becomes:
\[
x^2 + \psi^2 \geq \frac{1}{2} (x^2 + 2x\psi + \psi^2)
\]

### Step 2: Distribute the \(\frac{1}{2}\) on the Right Side
\[
x^2 + \psi^2 \geq \frac{1}{2}x^2 + x\psi + \frac{1}{2}\psi^2
\]

### Step 3: Rearrange Terms
Subtract \(\frac{1}{2}x^2\) and \(\frac{1}{2}\psi^2\) from both sides:
\[
x^2 - \frac{1}{2}x^2 + \psi^2 - \frac{1}{2}\psi^2 \geq x\psi
\]

Simplify the terms:
\[
\frac{1}{2}x^2 + \frac{1}{2}\psi^2 \geq x\psi
\]

### Step 4: Recognize as AM-GM Inequality
The inequality \(\frac{1}{2}x^2 + \frac{1}{2}\psi^2 \geq x\psi\) resembles the form of the Arithmetic Mean-Geometric Mean (AM-GM) inequality, which states that the arithmetic mean of non-negative numbers is always greater than or equal to their geometric mean.

Since \( \frac{1}{2}x^2 + \frac{1}{2}\psi^2 \geq x\psi \) holds by AM-GM, the original inequality is indeed valid for all real values of \(x\) and \(\psi\).

Would you like a deeper explanation of any step, or do you have further questions?

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Here are 5 related questions to deepen understanding:

1. How does the Arithmetic Mean-Geometric Mean (AM-GM) inequality apply to real numbers?
2. Can we solve for specific values of \(x\) and \(\psi\) where the inequality becomes equality?
3. How would the inequality change if the coefficients were different (e.g., \(\frac{1}{3}\) instead of \(\frac{1}{2}\))?
4. What other inequalities are commonly solved using AM-GM?
5. Can we interpret this inequality geometrically?

**Tip:** For inequalities, expanding terms carefully and using known inequalities like AM-GM can often simplify the solution process.

Prove the inequality x^2 + ψ^2 ≥ 1/2 (x + ψ)^2.

Explore the proof of the inequality x^2 + ψ^2 ≥ 1/2 (x + ψ)^2 using the Arithmetic Mean-Geometric Mean (AM-GM) inequality. This problem demonstrates algebraic expansion and simplification techniques for high school students, suitable for Grades 11-12.

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