Navjyot Kalra
Last Activity: 10 Years ago
Sol. We know that for sides a, b, c of a ∆
(a – b)2 ≥ 0
⇒ a2 + b2 ≥ 2ab . . . . . . . . . . . . . . . . (1)
Similarly b2 + c2 ≥ 2bc . . . . . . . . . . . . . . . .(2)
C2 + a2 ≥ 2ca . . . . . . . . . . . . . . . . .(3)
Adding the three inequations, we get
2(a2 + b2 + c2 ) ≥ 2(ab + bc + ca)
⇒ a2 + b2 + c2 ≥ ab + bc + ca
Adding 2 (ab + bc + ca) to both sides, we get
(a + b + c)2 ≥ 3(ab + bc + ca)
Or 3 (ab + bc + ca) ≤ (a + b + c)2 . . . . . . . . . . . (A)
Also c < a + b ( triangle inequality)
⇒ c2 < ac + bc . . . . . . . . . . . . . . .(4)
Similarly b2 < ab + bc . . . . . . . . . . . . . . (5)
a2 < ab + ca . . . . . . . . . . . . . . . . (6)
Adding (4), (5) and (6), we get a2 + b2 + c2 < 2(ab + bc + ca)
Adding 2 (ab + bc + ca) to both sides, we get
⇒ (a + b + c)2 < 4 (ab + bc + ca) . . . . . . . . . . (B)
Combining (A) and (B), we get
3(ab + bc + ca) ≤ (a + b + c)2 < 4(ab + bc + ca)
First two expressions will be equal for a = b = c.
