To prove the inequality:

[ \forall n \in \mathbb{N}^*, \forall (x, y) \in (\mathbb{R}^+)^2, \sqrt[n]{x + y} \leq \sqrt[n]{x} + \sqrt[n]{y}, ]

we can proceed using the following steps.

1. Define the Functions

Let $f_n(x) = \sqrt[n]{x}$, where $n \in \mathbb{N}^*$ and $x \in \mathbb{R}^+$. The inequality we want to prove is:

$f_n(x + y) \leq f_n(x) + f_n(y).$

2. Apply Minkowski's Inequality

Minkowski's inequality in $L^p$ space (for $p = \frac{1}{n}$, where $n > 0$) states that for any non-negative $a_1, a_2, \dots, a_m$ and $b_1, b_2, \dots, b_m$, the following inequality holds:

$\left(\sum_{i=1}^m (a_i + b_i)^p \right)^{\frac{1}{p}} \leq \left(\sum_{i=1}^m a_i^p \right)^{\frac{1}{p}} + \left(\sum_{i=1}^m b_i^p \right)^{\frac{1}{p}}.$

For $m = 2$ and $p = \frac{1}{n}$, consider $a_1 = x^{\frac{1}{n}}$ and $b_1 = y^{\frac{1}{n}}$. Then:

$(x^{\frac{1}{n}} + y^{\frac{1}{n}})^n \geq x + y.$

Taking the $n$-th root on both sides:

[ \sqrt[n]{x + y} \leq \sqrt[n]{x} + \sqrt[n]{y}. ]

3. Consider the Equality Case

Equality holds if and only if $x = 0$, $y = 0$, or $x^{\frac{1}{n}}$ and $y^{\frac{1}{n}}$ are proportional.

Conclusion

Thus, the inequality is proven:

[ \forall n \in \mathbb{N}^*, \forall (x, y) \in (\mathbb{R}^+)^2, \sqrt[n]{x + y} \leq \sqrt[n]{x} + \sqrt[n]{y}. ]

This concludes the proof.

Would you like further details or have any questions?

Here are some related questions to explore:

1. How can you apply Minkowski's inequality to higher dimensions?
2. What are other important inequalities in $L^p$ spaces?
3. How does Minkowski's inequality relate to the triangle inequality?
4. What is the significance of the equality condition in Minkowski's inequality?
5. How does this proof change if $x$ and $y$ can be negative?

Tip: Minkowski's inequality generalizes the triangle inequality for vector spaces, offering insights into the geometry of $L^p$ spaces.