ATQ, d + e > a + b ...(i) c + d > a + e ...(ii) a + b = c + d ...(iii) b + d = 2a ... (iv) (i)+(ii) d + e + c + d > a + b + a + e using (iii) e + d > a + e ∴ d > a Again, b + d = a + a (Given) and d > a' ∴ a > b (To hold equality) d + e > a + b ...(i) d + e > c + d using (iii) e > c Again from (ii) c + d > a + e a + b > a + e ⇒ b > e d > a > b > e > c