ThmDex – An index of mathematical definitions, results, and conjectures.
Result R1959 on D3042: Conjugate exponents
Young's inequality
Formulation 0
Let $x_1, \ldots, x_N \in [0, \infty)$ each be an D4767: Unsigned real number.
Let $\lambda_1, \ldots, \lambda_N \in (0, \infty)$ each be a D5407: Positive real number such that
(i) \begin{equation} \sum_{n = 1}^N \frac{1}{\lambda_n} = 1 \end{equation}
Then
(1) \begin{equation} \prod_{n = 1}^N x_n \leq \sum_{n = 1}^N \frac{1}{\lambda_n} x^{\lambda_n}_n \end{equation}
(2) \begin{equation} \prod_{n = 1}^N x_n = \sum_{n = 1}^N \frac{1}{\lambda_n} x^{\lambda_n}_n \quad \iff \quad x^{\lambda_1}_1 = x^{\lambda_2}_2 = \cdots = x^{\lambda_N}_N \end{equation}
Proofs
Proof 0
Let $x_1, \ldots, x_N \in [0, \infty)$ each be an D4767: Unsigned real number.
Let $\lambda_1, \ldots, \lambda_N \in (0, \infty)$ each be a D5407: Positive real number such that
(i) \begin{equation} \sum_{n = 1}^N \frac{1}{\lambda_n} = 1 \end{equation}
Applying R1557: Weighted real AM-GM inequality to the real numbers $y_n : = x^{\lambda_n}_n$ with weights $\theta_n : = 1 / \lambda_n$, we have \begin{equation} \prod_{n = 1}^N x_n = \prod_{n = 1}^N (x^{\lambda_n}_n)^{1 / \lambda_n} = \prod_{n = 1}^N y_n^{\theta_n} \leq \sum_{n = 1}^N \theta_n y_n = \sum_{n = 1}^N \frac{1}{\lambda_n} x^{\lambda_n}_n \end{equation} This establishes the first claim. The second claim follows from the equality condition of the same result. $\square$