Cauchy-Schwarz-like Inequality

Question

Let $a_i, b_i >0$ for all $n$ and $0 \le \lambda \le 1$

Is the following result true for all $n$?

$$ \sum^n_{i=0} a^\lambda_i b^{1-\lambda}_i \le \left( \sum^n_{i=0} a_i \right) ^\lambda \left( \sum^n_{i=0} b_i\right) ^{1-\lambda} $$

This is trivial for $n=1$, but I am getting stuck on the inductive step.

Answer 1

Straightforward from Hölder’s inequality in Euclidean spaces:

$$\|a\cdot b\|_1 \leq \|a\|_p \|b\|_q$$ for any $p,q\in (1,\infty) : 1/p+1/q=1 $, and for any $a,b \in \mathbb{R}^n$.

Let $ 1/p = \lambda $ and $1/q = 1 - \lambda $ when $ \lambda \in (0,1)$ since the other case when $\lambda = 0$ or $1$ is trivial. Then let $a= \tilde{a}^\lambda, b= \tilde{b}^{(1-\lambda)}$ and we are done

Answer 2

Step 1: It is easy (e.g., by taking the derivative) to show that $x^\lambda \le \lambda x + 1 - \lambda \ \forall x > 0$.

Step 2: Replacing $x$ in Step 1 with $\frac{x}{y}$ we obtain $x^\lambda y^{1-\lambda} \le \lambda x + (1 - \lambda)y \quad \forall x,y > 0.$

Step 3: Let $A = \sum_i a_i, B = \sum_i b_i$. Applying the inequality in Step 2 for each $i$ with $x = \frac{a_i}{A},y = \frac{b_i}{B}$. Summing up the resulting $n+1$ inequalities we obtain the result.