To prove that $|a + b| \leq \sqrt{2}$ given $a^2 + b^2 = 1$, let's proceed with the following steps:

Step 1: Express the sum $|a + b|$ in terms of $a$ and $b$.

The absolute value of the sum is given by: $|a + b| = \sqrt{(a + b)^2}$ Expanding the square: $(a + b)^2 = a^2 + 2ab + b^2$ Using the given condition $a^2 + b^2 = 1$: $(a + b)^2 = 1 + 2ab$ Thus, we have: $|a + b| = \sqrt{1 + 2ab}$

Step 2: Analyze the maximum possible value of $|a + b|$.

The expression $|a + b| = \sqrt{1 + 2ab}$ is maximized when $ab$ is maximized. Since $a^2 + b^2 = 1$, by the Cauchy-Schwarz inequality (or using the fact that $ab$ is maximized when $a = b$), the maximum value of $ab$ is $\frac{1}{2}$.

Substituting $ab = \frac{1}{2}$ into the expression for $|a + b|$: $|a + b| = \sqrt{1 + 2 \times \frac{1}{2}} = \sqrt{1 + 1} = \sqrt{2}$

Conclusion:

Since the expression $\sqrt{1 + 2ab}$ achieves its maximum value when $ab$ is at its maximum, the maximum value of $|a + b|$ is $\sqrt{2}$. Therefore: $|a + b| \leq \sqrt{2}$

This completes the proof that $|a + b| \leq \sqrt{2}$ given $a^2 + b^2 = 1$.

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Do you want more details or have any questions?

Here are 5 related questions that might interest you:

1. How does the Cauchy-Schwarz inequality relate to maximizing $ab$?
2. Can we generalize the result for $a^2 + b^2 = r^2$ instead of $1$?
3. What are the geometric interpretations of $a^2 + b^2 = 1$ and $|a + b|$?
4. How does this result apply to vectors in two-dimensional space?
5. What happens if $a^2 + b^2 \neq 1$?

Tip: When dealing with inequalities involving absolute values, consider squaring both sides to simplify the expression, but remember to check for extraneous solutions!