Why use “b < a ? a : b” instead of “a < b ? b : a” to implement max template?

std::max(a, b) is indeed specified to return a when the two are equivalent.

That's considered a mistake by Stepanov and others because it breaks the useful property that given a and b, you can always sort them with {min(a, b), max(a, b)}; for that, you'd want max(a, b) to return b when the arguments are equivalent.

This answer explains why the given code is wrong from a C++ standard point-of-view, but it is out of context.

See @T.C.'s answer for a contextual explanation.

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The standard defines std::max(a, b) as follows [alg.min.max] (emphasis is mine):

template<class T> constexpr const T& max(const T& a, const T& b);

Requires: Type T is LessThanComparable (Table 18).

Returns: The larger value.

Remarks: Returns the first argument when the arguments are equivalent.

Equivalent here means that !(a < b) && !(b < a) is true [alg.sorting#7].

In particular, if a and b are equivalent, both a < b and b < a are false, so the value on the right of : will be returned in the conditional operator, so a has to be on the right, so:

a < b ? b : a

...seems to be the correct answer. This is the version used by libstdc++ and libc++.

So the information in your quote seems wrong according to the current standard, but the context in which it is defined might be different.