The image contains a mathematical inequality that appears to be related to the Taylor series. Here's the inequality I extracted:

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

It also includes the requirement to prove it using the Taylor expansion.

Proof using Taylor Series:

Step 1: Expand $\sqrt{1 + 2x}$ using Taylor series around $x = 0$

The Taylor series expansion for $\sqrt{1 + u}$ (where $u = 2x$) around $u = 0$ is given by: $\sqrt{1 + u} = 1 + \frac{u}{2} - \frac{u^2}{8} + \frac{u^3}{16} + \dots$

Substitute $u = 2x$: $\sqrt{1 + 2x} = 1 + \frac{2x}{2} - \frac{(2x)^2}{8} + \frac{(2x)^3}{16} + \dots$

Simplify: $\sqrt{1 + 2x} = 1 + x - \frac{x^2}{2} + \frac{x^3}{2} + \dots$

Step 2: Compare the series with the given inequality

The given inequality is: $\sqrt{1 + 2x} \leq 1 + x - \frac{x^2}{2} + \frac{x^3}{2}.$

From the Taylor series expansion, the equality holds if we only consider terms up to $x^3$. For higher-order terms ($x^4$ and beyond), they will always decrease the value of $\sqrt{1 + 2x}$, making the inequality valid.

Conclusion

The inequality is proven using the Taylor expansion, as truncating the Taylor series up to $x^3$ leads to the exact form of the inequality, and higher-order terms confirm that $\sqrt{1 + 2x}$ remains less than or equal to the right-hand side.

Let me know if you'd like a more detailed breakdown or further clarification!

Expanding your understanding

Here are 5 related questions you might find useful:

1. What is the general formula for the Taylor expansion of $\sqrt{1 + u}$?
2. How does the error term in a Taylor approximation affect the inequality?
3. Why does the series expansion stop at $x^3$ in this proof?
4. Can we use Taylor series to prove other approximations like $\ln(1 + x)$?
5. How does the radius of convergence of a Taylor series impact such proofs?

Tip: Always verify the higher-order terms in a Taylor expansion when dealing with inequalities, as they determine the direction of the approximation!