ThmDex – An index of mathematical definitions, results, and conjectures.
Result R5189 on D3042: Conjugate exponents
Subresult of R5188:
Formulation 0
Let $x_1, y_1, \ldots, x_N, y_N \in [0, \infty)$ each be an D4767: Unsigned real number.
Let $\lambda_1, \ldots, \lambda_N \in [0, \infty)$ each be an D4767: Unsigned real number such that
(i) \begin{equation} \sum_{n = 1}^N \lambda_n = 1 \end{equation}
Let $p, q, \alpha \in (0, \infty)$ each be a D5407: Positive real number such that
(i) \begin{equation} \frac{1}{p} + \frac{1}{q} = 1 \end{equation}
Then
(1) \begin{equation} \sum_{n = 1}^N \lambda_n x_n y_n \leq \frac{\alpha^p}{p} \sum_{n = 1}^N \lambda_n x^p_n + \frac{1}{q \alpha^q} \sum_{n = 1}^N \lambda_n y^q_n \end{equation}
(2) \begin{equation} \sum_{n = 1}^N \lambda_n x_n y_n = \frac{\alpha^p}{p} \sum_{n = 1}^N \lambda_n x^p_n + \frac{1}{q \alpha^q} \sum_{n = 1}^N \lambda_n y^q_n \quad \iff \quad \forall \, 1 \leq n \leq N : \frac{\alpha^p}{p} x^p_n = \frac{1}{q \alpha^q} y^q_n \end{equation}
Proofs
Proof 0
Let $x_1, y_1, \ldots, x_N, y_N \in [0, \infty)$ each be an D4767: Unsigned real number.
Let $\lambda_1, \ldots, \lambda_N \in [0, \infty)$ each be an D4767: Unsigned real number such that
(i) \begin{equation} \sum_{n = 1}^N \lambda_n = 1 \end{equation}
Let $p, q, \alpha \in (0, \infty)$ each be a D5407: Positive real number such that
(i) \begin{equation} \frac{1}{p} + \frac{1}{q} = 1 \end{equation}
Let $n = 1, \ldots, N$. Using result R5188: and multiplying both sides by $\lambda_n$, we have the inequality \begin{equation} \lambda_n x_n y_n \leq \frac{\alpha^p}{p} \lambda_n x^p_n + \frac{1}{q \alpha^q} \lambda_n y^q_n \end{equation} with equality if and only if $\frac{\alpha^p}{p} x^p_n = \frac{1}{q \alpha^q} y^q_n$. Summing both sides over $n = 1, \ldots, N$ now yields the claim. $\square$