Example 6.2.1. Suppose that \(x, y, z\) are three positive real numbers verifying \(x + y + z = 3\). Find the minimum of the expression
\[
x^{4} + 2 y^{4} + 3 z^{4}
\]
Solution. Let $a, b, c$ be three positive real numbers such that $a+b+c=3$. According to Hölder inequality, we obtain
$$\left(x^{4}+2 y^{4}+3 z^{4}\right)\left(a^{4}+2 b^{4}+3 c^{4}\right)^{3} \geq\left(a^{3} x+2 b^{3} y+3 c^{3} z\right)^{4}(\star)$$

We will choose $a, b, c$ such that $a^{3}=2 b^{3}=3 c^{3}=k^{3}$, and if then
$$x^{4}+2 y^{4}+3 z^{4} \geq \frac{k^{12}(x+y+z)^{4}}{\left(a^{4}+2 b^{4}+3 c^{4}\right)^{3}}=\frac{\left(3 k^{3}\right)^{4}}{\left(a^{4}+2 b^{4}+3 c^{4}\right)^{3}}(\star \star)$$

The equality in ( $\star$ ) happens when $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$. Because $x+y+z=a+b+c=3$, we get $a=x, b=y, c=z$. We find that $k=\frac{3}{1+\sqrt[3]{2}+\sqrt[3]{3}}$ and $a=k, b=\sqrt[3]{2} k, c=\sqrt[3]{3} k$. The minimum of $x^{4}+2 y^{4}+3 z^{4}$ is given by $(\star \star)$.