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Mathematics
Problem
Prove Cauchy-Schwartz inequality
(
a
⋅
b
)
2
≤
∣
a
∣
2
⋅
∣
b
∣
2
(\mathbf{a} \cdot \mathbf{b})^2 \le |\mathbf{a}|^2 \cdot |\mathbf{b}|^2
(
a
⋅
b
)
2
≤
∣
a
∣
2
⋅
∣
b
∣
2
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