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Exercise 1.1.5 Let $n \ge 1$, and let $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ be real numbers. Verify the identity $\left(\sum_{i=1}^n a_i b_i \right)^2 + \frac{1}{2} \sum_{i=1}^n \sum_{j=1}^n (a_i b_j - a_j b_i)^2 = \left(\sum_{i=1}^n a_i^2 \right) \left(\sum_{j=1}^n b_j^2 \right)$. and conclude the Cauchy-Schwarz inequality $\left| \sum_{i=1}^n a_i b_i \right| \le \left( \sum_{i=1}^n a_i^2 \right)^{1/2} \left( \sum_{j=1}^n b_j^2 \right)^{1/2}$. (1.3) Then use the Cauchy-Schwarz inequality to prove the triangle inequality $\left( \sum_{i=1}^n (a_i + b_i)^2 \right)^{1/2} \le \left( \sum_{i=1}^n a_i^2 \right)^{1/2} + \left( \sum_{j=1}^n b_j^2 \right)^{1/2}$.

          Exercise 1.1.5 Let $n \ge 1$, and let $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ be real numbers.
Verify the identity
$\left(\sum_{i=1}^n a_i b_i \right)^2 + \frac{1}{2} \sum_{i=1}^n \sum_{j=1}^n (a_i b_j - a_j b_i)^2 = \left(\sum_{i=1}^n a_i^2 \right) \left(\sum_{j=1}^n b_j^2 \right)$.
and conclude the Cauchy-Schwarz inequality
$\left| \sum_{i=1}^n a_i b_i \right| \le \left( \sum_{i=1}^n a_i^2 \right)^{1/2} \left( \sum_{j=1}^n b_j^2 \right)^{1/2}$.
(1.3)
Then use the Cauchy-Schwarz inequality to prove the triangle inequality
$\left( \sum_{i=1}^n (a_i + b_i)^2 \right)^{1/2} \le \left( \sum_{i=1}^n a_i^2 \right)^{1/2} + \left( \sum_{j=1}^n b_j^2 \right)^{1/2}$.

Exercise 1.1.5 Let n ≥ 1, and let a1, a2, …, an and b1, b2, …, bn be real numbers.
Verify the identity
(∑i=1^n ai bi )^2 + (1)/(2)∑i=1^n ∑j=1^n (ai bj - aj bi)^2 = (∑i=1^n ai^2 ) (∑j=1^n bj^2 ).
and conclude the Cauchy-Schwarz inequality
| ∑i=1^n ai bi | ≤( ∑i=1^n ai^2 )^1/2( ∑j=1^n bj^2 )^1/2.
(1.3)
Then use the Cauchy-Schwarz inequality to prove the triangle inequality
( ∑i=1^n (ai + bi)^2 )^1/2≤( ∑i=1^n ai^2 )^1/2 + ( ∑j=1^n bj^2 )^1/2.

Added by Richard S.

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