Given that P<Q<R<S,

Can P = 3?  No, because 1/3 + 1/4 + 1/5 + 1/6 < 1

So if the largest fraction were 1/3, it would be too small; or 3 is too big.

Therefore P = 2.

If P=2, then 1/Q + 1/R + 1/S = 1/2

What values can Q take?

Q cannot be 2 since the remaining 2 fractions would have to be 0.

Q cannot be 6 since 1/2 + 1/6 + 1/7 + 1/8 < 1.

Therefore Q is in {3,4,5}

Given P and Q, find possible R values such that:

1/P + 1/Q + 1/R < 1, but 1/P + 1/Q + 1/R + 1/(R+1) > 1

For (P,Q) = (2,3) R must be greater than 6 but smaller than 12

For (P,Q) = (2,4) R must be greater than 4 but smaller than 8

For (P,Q) = (2,5) R must be greater than 5 but smaller than 7

So far, there are only 9 possible values for (P,Q,R):

(2,3,7)(2,3,8)(2,3,9)(2,3,10)(2,3,11)

(2,4,5)(2,4,6)(2,4,7)

(2,5,6)

Next calculate what real value of Q satisfies 1/(1 - (1/P + 1/Q + 1/R))

We get only 6 solutions with integer values for Q:

(2,3,7,42)

(2,3,8,24)

(2,3,9,18)

(2,3,10,15)

(2,3,11,13.2)   rejected

(2,4,5,20)

(2,4,6,12)

(2,4,7,9.33333) rejected

(2,5,6,7.5)     rejected

----------

This matches the output of the computer program I ran, initially letting all 4 variables vary from 2 to 200, though later I trimmed down the range of values to be tested after determining limits for P and Q.  Also I multiplied both sides of the equation by PQRS to avoid rounding errors associated with division.

Program Output:

(2,3,7,42)

(2,3,8,24)

(2,3,9,18)

(2,3,10,15)

(2,4,5,20)

(2,4,6,12)

"""

p=2

big = 200

for q in range(3,6):

    for r in range(q+1,big):

        for s in range(r+1,big):

            if q*r*s + p*r*s + p*q*s + p*q*r  == p*q*r*s:

                print('({},{},{},{})'.format(p,q,r,s))