Как заменить элемент в списке по индексу и сдвинуть наименьшее на начальную позицию?

def f(lst, x):
    for i in range(len(lst)):
        if lst[i] < x:
            lst[i] = x
            break
    lst.sort()
    return lst

Воспользуйтесь модулем bisect:

This module provides support for maintaining a list in sorted order without having to sort the list after each insertion. For long lists of items with expensive comparison operations, this can be an improvement over the more common approach. The module is called bisect because it uses a basic bisection algorithm to do its work. The source code may be most useful as a working example of the algorithm (the boundary conditions are already right!).

решение:

import bisect

def fun(lst, x):
    bisect.insort(lst, x)
    return lst[1:]

тесты:

In [57]: res = fun(chips, my_chips)

In [58]: res
Out[58]: [700, 750, 750, 900]

In [59]: res2 = fun([500, 900, 1200], 3000)

In [60]: res2
Out[60]: [900, 1200, 3000]

Вот так вот для непустых списков:

import itertools


def pairwise(iterable):
    a, b = itertools.tee(iterable)
    next(b, None)
    return zip(a, b)


def modify_list(l, val):
    i = 0

    for a, b in pairwise(itertools.takewhile(lambda x: x < val, l)):
        l[i], l[i + 1] = b, a
        i += 1

    if l[i] < val:
        l[i] = val


chips = [400, 700, 900]
my_chips = 750

modify_list(chips, my_chips)
print(chips)  # [700, 750, 900]

chips = [500, 900, 1200]
my_chips = 3000

modify_list(chips, my_chips)
print(chips)  # [900, 1200, 3000]

chips = [500]
my_chips = 3000

modify_list(chips, my_chips)
print(chips)  # [3000]

chips = [4000]
my_chips = 3000

modify_list(chips, my_chips)
print(chips)  # [4000]