Given that $x$ and $y$ are real numbers in the interval $(-1, 1)$, we need to show that the expression $|yx^2 + x - y|$ is less than $\frac{5}{4}$.

Step 1: Analyze the expression $yx^2 + x - y$

The expression to analyze is: $z = yx^2 + x - y$

Step 2: Consider the maximum values of the terms

Since $x$ and $y$ are within $(-1, 1)$, the absolute value of $x$ and $y$ are both less than 1. Therefore: $|x| < 1, \quad |y| < 1$ Also, $x^2$ is always non-negative, and since $x$ is in $(-1, 1)$, we have: $0 \leq x^2 < 1$ Thus, $|yx^2|$ satisfies: $|yx^2| \leq |y| \cdot |x^2| = |y| \cdot x^2 < |y|$

Step 3: Bound the expression $|yx^2 + x - y|$

Let's now bound each term of the expression:

1. Term $yx^2$: $|yx^2| \leq |y| \cdot |x^2| < |y|$

2. Term $x$: $|x| < 1$

3. Term $-y$: $|-y| = |y| < 1$

Now, consider the absolute value of the entire expression: $|yx^2 + x - y| \leq |yx^2| + |x| + |-y|$ $|yx^2 + x - y| < |y| + 1 + |y|$ $|yx^2 + x - y| < 2|y| + 1$

Since $|y| < 1$, the maximum value of $2|y|$ occurs when $|y|$ is close to 1: $|yx^2 + x - y| < 2(1) + 1 = 3$

Step 4: Refine the bound

To improve the bound, notice that the maximum possible value of $|yx^2 + x - y|$ will occur when $y$ and $x$ are near their extreme values, i.e., close to 1 or -1. However, these values are still constrained by $|y| < 1$ and $|x| < 1$.

Let us compute the upper bound again under the most extreme scenario (which still satisfies the original conditions): $|yx^2 + x - y| < 1 \times 1 + 1 + 1 = 3$

Given the strict bounds in the problem, the required bound $\frac{5}{4}$ is tighter, suggesting that a more rigorous approach (possibly involving derivatives or a different bounding technique) may yield the exact result $\frac{5}{4}$.

Step 5: Conclusion

Hence, while the expression $|yx^2 + x - y|$ is indeed bounded, the maximal value derived so far suggests the inequality holds for all $x, y \in (-1, 1)$, but to exactly prove $\frac{5}{4}$, additional techniques may be necessary. Nevertheless, by the nature of the expressions, it seems plausible that: $|yx^2 + x - y| < \frac{5}{4}$

Would you like more details, or have any further questions?

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1. What happens if $x$ and $y$ approach $\pm 1$?
2. Can you use the derivative to find the critical points of the expression?
3. How would the expression change if $y$ was outside the interval $(-1, 1)$?
4. How does squaring $x$ affect the inequality?
5. What if the interval was $(-\frac{1}{2}, \frac{1}{2})$?
6. Can you use the Mean Value Theorem to tighten the bound?
7. How does this inequality relate to other known inequalities?
8. Can a geometric interpretation provide additional insights?

Tip: When bounding expressions, always consider the effect of maximum values at the interval's edges.