We have
$a+b+c = 1$
So $1+a + b+c = 2$
Or $1+a = 2 - b-c = (1-b)+(1-c)$
because (1-b) and (1-c) are positive using AM GM inequality we have
$(1-b)(a-c) >= 2\sqrt{(1-b)(1-c)}$
or $(1+a) >= 2\sqrt{(1-b)(1-c)}\cdots(1)$
similarly
$(1+b) >= 2\sqrt{(1-c)(1-a)}\cdots(2)$
and
$(1+c) >= 2\sqrt{(1-a)(1-b)}\cdots(3)$
Multiply above 3 equations we get
$(1+a)(1+b)(1+c) >= 8(1-a)(1-b)(1-c)$
Let there be finite number of prime numbers of the form 4n+ 3 the number of numbers by k with $p_k = 4 q_k +3$
Now consider the number $4p_1p_2\cdots p_k -1 $
The number above is of the form 4n+3.
If the number is prime then we have found a prime number above the largest prime and we are done.
If it is not a prime number it has got some prime factors.
All the prime factors cannot be of the 4n+1 because in that case product shall be of the from 4n+ 1
So it has got a prime factor of the form 4n+3 above $p_k$. which is again a conradiction.
Hence there are infinite prime numbers of the form 4n+3
