In a inner product space show that the following inequality holds.
$\|x-z\|.\|y-t\|\leq \|x-y\|.\|z-t\|+\|y-z\|.\|x-t\|$
I am stuck in proving this inequality
2026-04-04 05:22:45.1775280165
An inequality for inner product space: $\|x-z\|.\|y-t\|\leq \|x-y\|.\|z-t\|+\|y-z\|.\|x-t\|$
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This is Ptolemy's inequality, in the (at most) $3$-dimensional vector space spanned by differences of the $4$ vectors, and by a linear transformation is equivalent to the case when the inner product is the standard one, $\sum x_i y_i$.
In $2$ dimensions equality holds when the points are on a circle, and proofs are not obvious, though not difficult once the statement is given. It looks like the formula is an identity without the $||$ signs, replacing products of numbers with inner products of vectors, and it would be nice to conclude by going from $A = B+C$ to $|A| \leq |B| + |C|$. But the inner products can be boosted from numbers to elements in a vector space, such as Sym^2$(V)$ or $V \otimes V$, where $V$ is the vector space under discussion, so this could be a legitimate argument. If those musings do not work out, search engines will locate proofs from the words "Ptolemy inequality".