If a, b, c are the sides of a Δ ABC, then show that \(\frac12<\frac{

1.	If a, b, c are the sides of a Δ ABC, then show that \(\frac12<\frac{ab+bc+ca}{a^2+b^2+c^2}\) ≤ 1.
Answer» We have a > 0, b > 0, c > 0, so (a – b)² + (b – c)² + (c – a)² > 0 ⇒ a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca > 0 ⇒ 2a² + 2b² + 2c² > 2ab + 2bc + 2ca ⇒ a² + b² + c² > ab + bc + ca ⇒ 1 > \(\frac{ab+bc+ca}{a^2+b^2+c^2}\) ⇒ \(\frac{ab+bc+ca}{a^2+b^2+c^2}\) < 1 …(i) Now for the other part let us using the triangle inequality, i.e., the sum of two sides of a triangle is greater than the third side, we have ∴ a + b > c, b + c > a, c + a > b. Also, a > 0, b > 0, c > 0 ⇒ a + b – c > 0, b + c – a > 0, c + a – b > 0 ⇒ a (b + c – a) + b (c + a – b) + c (a + b – c) > 0 ⇒ ab + ac – a² + bc + ba – b² + ca + cb – c² > 0 ⇒ 2 (ab + bc + ac) > a² + b² + c² ⇒ \(\frac{ab+bc+ca}{a^2+b^2+c^2}≥\frac12\) ....(ii) ∴ From (i) and (ii) \(\frac12<\frac{ab+bc+ca}{a^2+b^2+c^2}≥1.\)

If a, b, c are the sides of a Δ ABC, then show that \(\frac12<\frac{ab+bc+ca}{a^2+b^2+c^2}\) ≤ 1.

Answer»

We have a > 0, b > 0, c > 0, so

(a – b)² + (b – c)² + (c – a)² > 0

⇒ a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca > 0

⇒ 2a² + 2b² + 2c² > 2ab + 2bc + 2ca ⇒ a² + b² + c² > ab + bc + ca

⇒ 1 > \(\frac{ab+bc+ca}{a^2+b^2+c^2}\) ⇒ \(\frac{ab+bc+ca}{a^2+b^2+c^2}\) < 1 …(i)

Now for the other part let us using the triangle inequality,

i.e., the sum of two sides of a triangle is greater than the third side, we have

∴ a + b > c, b + c > a, c + a > b. Also, a > 0, b > 0, c > 0

⇒ a + b – c > 0, b + c – a > 0, c + a – b > 0

⇒ a (b + c – a) + b (c + a – b) + c (a + b – c) > 0

⇒ ab + ac – a² + bc + ba – b² + ca + cb – c² > 0

⇒ 2 (ab + bc + ac) > a² + b² + c² ⇒ \(\frac{ab+bc+ca}{a^2+b^2+c^2}≥\frac12\) ....(ii)