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

Question

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 1

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) 

\(\frac12<\frac{ab+bc+ca}{a^2+b^2+c^2}≥1.\)