see the counterexample.
Work Step by Step
Assume that $u=(1,1,1)$, then we have $$\langle u,u \rangle =u_1^2-u_2^2-u_3^2\Longrightarrow \langle (1,1,1),(1,1,1) \rangle =-1<0.$$ This shows that the function does not define an inner product.
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