Upper bound of the dot product of two real-valued vectors

alternative-proofinner-productslinear algebrasolution-verification

I am interested in proving the following:

$ \sum_{i=1}^n a_ib_i \leq \max \{\sum_{i=1}^n a^2_i,\sum_{i=1}^n b^2_i\}$, where $a_i,b_i \in \mathbb{R}$.

I came up with a geometric proof and a proof by contradiction but I feel like they don't give me much insight as to what really happens.

For the geometric proof I say that the collections of $a_i$ and $b_i$ can be seen as two real-valued vectors of $\mathbf{a},\mathbf{b}\in \mathbb{R}^n$ indexed by $i$.
Then, the original sum can be seen as the dot product of the vectors $\mathbf{a},\mathbf{b}$:

$\mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\|\cos\theta\,.$

We can then derive the following:

$ \|\mathbf{a}\|\|\mathbf{b}\|\cos\theta \leq \|\mathbf{a}\|\|\mathbf{b}\| \leq \max \{\|\mathbf{a}\|^2,\|\mathbf{b}\|^2\}\,.$

Assuming that $\|\mathbf{a}\|$ denotes the Euclidian norm of $\mathbf{a}$ we get:

$ \sum_{i=1}^n a_ib_i \leq \max \{\sum_{i=1}^n a^2_i,\sum_{i=1}^n b^2_i\}$. QED

For the proof by contradiction I just assume that if $ \sum_{i=1}^n a_ib_i > \max \{\sum_{i=1}^n a^2_i,\sum_{i=1}^n b^2_i\}$ then it implies that there exists a $a_i,b_i$ pair such that $a_ib_i > \max\{a^2_i,b^2_i\}$ but that is a contradiction. QED

I am not sure that these proofs are correct, although I am pretty convinced that the geometric one is.
I feel like the geometric proof gives me no insight because I don't really understand the concept of a norm in higher dimensions.

Is there a way to prove this algebraically?
I also tried proving it by induction but it lead me nowhere.

Best Answer

Let $a,b\in\mathbb{R}^n$. Then $$ 0\le \| a-b \|^2 = \| a \|^2-2a\cdot b+\| b\|^2, $$ so we have $$ a\cdot b \le \frac{\|a\|^2+\|b\|^2}{2}\le \max\{\|a\|^2,\|b\|^2\}. $$

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