Minimum value a+b+c+d

Let's start by getting an estimate of our minimum value $a+b+c+d$ by solving for the minimum value of $a+b+c+d$ when $a,b,c,$ and $d$ are positive real numbers.

In that case we'd get $a=b=c=d=\sqrt[4]{10!}\approx 43.6456$ with $a+b+c+d=4\sqrt[4]{10!}\approx 174.583$ . Therefore when $a,b,c,$ and $d$ are positive integers $a+b+c+d\ge 175$ .

Knowing that we'd ideally want values for $a,b,c,$ and $d$ that are near 44, I calculated the prime factors of 10! and then regrouped them with this in mind.

$10!= 1*2*3*4*5*6*7*8*9*10 =$ $2*3*(2*2)*5*(2*3)*7*(2*2*2)*(3*3)*(2*5)=$ $(2^3*5)*(2*3*7)*(3*3*5)*(2^4*3)=$ $40*42*45*48$

$40+42+45+48=175$

Having an solution where $a+b+c+d=175$ and knowing that $a+b+c+d\ge 175$ brings us to the conclusion that the minimum value of $a+b+c+d$ is $175$ .