a^2 9^ +b^2 27^ +c^2 63^ +d^2 81^ Let a, b, c and d be real numbers satisfying a+b+c+d=5 . If the minimum value of the expression above is equal to x y , where x and y are co-prime positive integers, find x+y

Step-by-step solution

(a^2 9^ +d^2 9^ )+ (b^2 27^ +c^2 27^ ) ...(1) Given a+b+d=5 Let a+d=k₁ aligned b+c & =k₂ Let k₁+k₂ & =5 E₁ & =a^2 9^ +d^2 9^ E₁ & =a^2 9^ + (k₁-a )^2 9^ ...(2) E₁ & =a^2 ( 9^ + 9^ )-a k₁ 9^ +k₁^2 9^ aligned This is quadratic in a whose minimum value is obtained at x= -B 2 A i.e., a= 2 k₁ 9^ 2 ( 9^ + 9^ ) =k₁ ^2 9^ Let aligned E₂ & =b^2 27^ +c^2 27^ b+c & =k₂ E₂ & =b^2 27^ + (k^2-b )^2 27^ ...(3) aligned Make quadratic in bE₂ ~min is obtained b=k₂ ^2 27^ From Eqn. (2) aligned E_ 1 & =k₁^2 ^4 9^ 9^ + (k₁ ^2 9 )^2 9^