This problem is beyond IIT syllabus but only by a hair's breadth. In any case, this problem's solution is worth knowing because of an inequality employed that is very beautiful and useful. In fact, I have been looking for an opportunity to introduce you guys to it after I came to know of it recently and this is the perfect place. The proof is in two parts.

Here, I assume that p>q>r and a>b>c which is not given in the problem.

Part 1: Here is where we use that beautiful inequality. The inequality in simplest terms is this: Consider two sequences x₁>x₂>x₃ and y₁>y₂>y₃ of positive terms . You can constitute various sums x_iy_j such as say x₁y₂+x₂y₃+x₃y₁. The largest such sum is x₁y₁+x₂y₂+x₃y₃ and the least is x₁y₃+x₂y₂+x₃y₁. This is called the Rearrangement Inequality. It can be extended to any number of terms and any number of sequences.

 

We can employ this result to do away with p,q, and r in a very easy manner

 

p>q>r means p+q>p+r>q+r and hence 1/q+r>1/p+r>1/p+q.

 

Also a>b>c means a²>b²>c²

 

Hence a²/q+r>b²/p+r>c²/p+q  .....1

           p        >q       >r ..............2

 

Hence (pa²/q+r) + (qb²/p+r) + rc²/(p+q) > (qa²/q+r) + (rb²/p+r) + pc²/(p+q)

 

  and   (pa²/q+r) + (qb²/p+r) + rc²/(p+q) > (ra²/q+r) + (pb²/p+r) + qc²/(p+q)

 

Add the above two inequalities to get 2I > a²+b²+c².............I

where I =  (pa²/q+r) + (qb²/p+r) + rc²/(p+q)

 

Isnt it amazing how neatly we made p,q, and r vanish into thin air with this fantastic artifice. So now on to Part II.

 

Part II: This is a somewhat well known result:

a²+b²+c²>43S where S is the area of the triangle.

 

We know that a²+b²+c²>ab+bc+ca.

We also know that S = 0.5*ab*sinC. Hence ab = 2S/sinC

 

Hence ab+bc+ca = 2S(1/sinC+1/sinA+1/sinB)

 

Now sinA+SinB+SinC < 33/2 is a well known result

 

Now we use (sinA+SinB+sinC)*(1/sinA+1/sinB+1/sinC)>9 (by using AM-GM on the two expressions being multiplied). Actually this is a standard result in its own right that if x,y and z are positive reals, then (x+y+z)*(1/x+1/y+1/z)>9

 

Hence 1/sinA+1/sinB+1/sinC > 9/(sinA+sinB+sinc) > 9/(33/2 ) = 23

Thus ab+bc+ca = 2S(1/sinA+1/sinB+1/sinC) > 2S*23 = 4S3. ......II

 

From I and II, we get 2I>4S3. or I>2S3.

 

Thus (pa²/q+r) + (qb²/p+r) + rc²/(p+q) > 2S3

 
In Part II, wherever I used >, I actually meant >=.