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Interesting Inequality

Thing and a two

Augustin-Louis Cauchy

Problem 1 (Interesting Inequality) Let \(f: \R_4^+ \to \R\) be the function defined by \(\bm{z} \mapsto z_1^2 z_2 z_3 + z_2^2 z_3 z_4 + z_3^2 z_4 z_1 + z_4^2 z_1 z_2\). Show that \[ f(\bm{z}) \le \norm{\bm{z}}{4}^4 = z_1^4 + z_2^4 + z_3^4 + z_4^4. \]

Proof. Define the two vectors \[ \bm{u} = \bmat{z_1^2 & z_2^2 & z_3^2 & z_4^2}, \quad \bm{v} = \bmat{z_2 z_3 & z_3 z_4 & z_4 z_1 & z_1 z_2}. \] We can write \(f(\bm{z})\) as the inner product of these two vectors: \[ f(\bm{z}) = \bm{u} \cdot \bm{v}. \] By the Cauchy-Schwarz inequality, we have \[ (\bm{u} \cdot \bm{v})^2 \le \|\bm{u}\|^2 \|\bm{v}\|^2 = (z_1^4 + z_2^4 + z_3^4 + z_4^4)\left( (z_2 z_3)^2 + (z_3 z_4)^2 + (z_4 z_1)^2 + (z_1 z_2)^2 \right). \tag{1}\] Now, let us define the shifted version \(\bm{u}_s\) of the vector \(\bm{u}\) \[ \bm{u}_s = \bmat{z_2^2 & z_3^2 & z_4^2 & z_1^2}. \] Clearly \(\norm{\bm{u}_s}{} = \norm{\bm{u}}{}\). Apply Cauchy-Schwarz inequality once more, this time to the vectors \(\bm{u}\) and \(\bm{u}_s\). We have \[ \left( (z_2 z_3)^2 + (z_3 z_4)^2 + (z_4 z_1)^2 + (z_1 z_2)^2 \right) = \bm{u} \cdot \bm{u}_s \le \|\bm{u}\|^2 = z_1^4 + z_2^4 + z_3^4 + z_4^4. \tag{2}\] Substituting from Equation 2 into Equation 1, we obtain \[ (\bm{u} \cdot \bm{v})^2 \le (z_1^4 + z_2^4 + z_3^4 + z_4^4)^2 \; \Leftrightarrow f(\bm{z}) \le z_1^4 + z_2^4 + z_3^4 + z_4^4. \]

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