use AM≥GM

6r1+10r2+15r3+30r4 ≥4(27000*r1r2r3r4)^0.25=120 as r1r2r3r4=30

therefore 6r1=10r2=15r3=30r4 =120/4=30

hence r1=5 ;r2=3; r3=2; r4=1

c-b=-20 ; a=11

hence answer=400-121=279

This is $MY$ Solution. I don't know whether it is correct or not. I was just lucky.

$6r_1 + 10r_2 + 15r_3 + 30r_4 = 120$ In this we can see that there are four terms summing up to $120$ . Letting the four terms to be equal to each other, we get each term to be = $30$ . This implies that

$6r_1 = 30 \rightarrow r_1 = 5$

Similarly $r_2 = 3$ , $r_3 =2$ and $r_4 = 1$

So we have the roots $\rightarrow 1,2,3,5$

Using Vieta's Theorem, we will get $a = 11, b =41$ and $c = 61$

Thus $(c - b)^{2} - a^{2} \\= (61 - 41)^2 -11^2 \\= 20^2 - 11^2 \\= 400 - 121 \\= 279$

Here's a similar solution to what Aakash Khandelwal had written. Just want to write it up for clarity.

We can apply the AM - GM Inequality to the set of positive reals: { $6r_1, 10r_2, 15r_3, 30_4$ }. That will be:

$\frac{6r_1 + 10r_2 + 15r_3 + 30r_4}{4}$ $\ge$ $\sqrt[4]{6r_1 × 10r_2 × 15r_3 × 30r_4}$ $\Longleftrightarrow$ $\frac{120}{4}$ $\ge$ $\sqrt[4]{6 × 10 × 15 × 30 × 30}$ (since $r_1r_2r_3r_4 = 30$ ) $\Longleftrightarrow$ $30$ $\ge$ $30$

We have an equality! And yes, that only occurs when $6r_1 = 10r_2 = 15r_3 = 30r_4 = 30$ . So for that, we should get the positive real roots as said on the problem: $r_1 = 5, r_2 = 3, r_3 = 2$ , and $r_4 = 1$ . Well from Vieta's Identities, that should immediately gives us the corresponding values of $a, b$ , and $c$ . And so,

$a = 5 + 3 + 2 + 1 = 11$

$b = 5 × 3 + 5 × 2 + 5 × 1 + 3 × 2 + 3 × 1 + 2 × 1 = 41$

$c = 5 × 3 × 2 + 5 × 3 × 1 + 5 × 2 × 1 + 3 × 2 × 1 = 61$

Our desire value of $(c - b)^2 - a^2$ should be $(61 - 41)^2 - 11^2 = 279$ .