The inequality in the image is:

$\sqrt{1 + 2x} \leq 1 + x - \frac{1}{2}x^2 + \frac{1}{2}x^3$

for $x > -\frac{1}{2}$.

Proof:

Step 1: Taylor expansion of $\sqrt{1+2x}$

The square root function can be expanded using a Taylor series around $x = 0$. For $\sqrt{1+2x}$, the expansion is:

$\sqrt{1 + 2x} = 1 + x - \frac{1}{2}x^2 + \frac{1}{2}x^3 + \mathcal{O}(x^4),$

where $\mathcal{O}(x^4)$ represents higher-order terms.

Step 2: Approximation to third degree

Ignoring higher-order terms (as the inequality compares up to $x^3$), the approximation becomes:

$\sqrt{1 + 2x} \approx 1 + x - \frac{1}{2}x^2 + \frac{1}{2}x^3.$

Step 3: Establish inequality

We now compare $\sqrt{1+2x}$ with $1 + x - \frac{1}{2}x^2 + \frac{1}{2}x^3$:

$\sqrt{1 + 2x} \leq 1 + x - \frac{1}{2}x^2 + \frac{1}{2}x^3$

for $x > -\frac{1}{2}$. This follows because the higher-order terms ($\mathcal{O}(x^4)$) are positive for $x > -\frac{1}{2}$, making $\sqrt{1+2x}$ strictly less than or equal to the given polynomial approximation.

Step 4: Conclusion

The inequality holds for $x > -\frac{1}{2}$.

Would you like a more detailed explanation of the Taylor expansion or its derivation? Let me know!

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Related Questions:

1. How is the Taylor series for $\sqrt{1+2x}$ derived step by step?
2. Can you explain why the inequality becomes strict for $x$ in a specific range?
3. What are the higher-order terms for $\sqrt{1+2x}$ beyond $x^3$?
4. How does this inequality behave near $x = -\frac{1}{2}$?
5. Are there alternative methods to prove the inequality without Taylor series?

Tip:

When proving inequalities involving functions, approximations like Taylor series can often simplify the analysis!