If a, b, c are the lengths of the sides of a non-equilateral triangle,

1.	If a, b, c are the lengths of the sides of a non-equilateral triangle, then \(\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\) is(a) > \(\frac{s}{a+b+c}\)(b) > \(\frac{2s}{abc}\)(c) < \(\frac{2s}{abc}\) (d) None of these, where s = \(\frac{a+b+c}{2}\).
Answer» (b) > \(\frac{2s}{abc}\) a > 0, b > 0, c > 0 ⇒ ab > 0, bc > 0, ca > 0 ∴ \(\frac{(ab)^2+(bc)^2}{2}\) > \(\big((ab)^2(bc)^2\big)^{\frac12}\) ⇒ a²b² + b²c² > 2 (a²b⁴c²)^\(\frac12\) ⇒ a²b² + b²c² > 2 (abc).b …(i) Similarly, b²c² + c²a² > 2 (abc).c …(ii) c²a² + a²b² > 2(abc)a …(iii) (i) + (ii) + (iii) ⇒ 2(a²b² + b²c² + c²a²) > 2 abc (a + b + c) ⇒ a²b² + b²c² + c²a² > abc (a + b + c) Dividing both sides by a²b²c², we have ⇒ \(\frac{1}{c^2}+\frac{1}{a^2}+\frac{1}{b^2}\) > \(\frac{a+b+c}{abc}\) ⇒ \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\) > \(\frac{2s}{abc}\) ⇒ \(\frac{2s}{abc}\) < \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\).

If a, b, c are the lengths of the sides of a non-equilateral triangle, then \(\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\) is(a) > \(\frac{s}{a+b+c}\)(b) > \(\frac{2s}{abc}\)(c) < \(\frac{2s}{abc}\) (d) None of these, where s = \(\frac{a+b+c}{2}\).

Answer»

(b) > \(\frac{2s}{abc}\)

a > 0, b > 0, c > 0 ⇒ ab > 0, bc > 0, ca > 0

∴ \(\frac{(ab)^2+(bc)^2}{2}\) > \(\big((ab)^2(bc)^2\big)^{\frac12}\)

⇒ a²b² + b²c² > 2 (a²b⁴c²)^\(\frac12\)

⇒ a²b² + b²c² > 2 (abc).b …(i)

Similarly, b²c² + c²a² > 2 (abc).c …(ii)

c²a² + a²b² > 2(abc)a …(iii)

(i) + (ii) + (iii)

⇒ 2(a²b² + b²c² + c²a²) > 2 abc (a + b + c)

⇒ a²b² + b²c² + c²a² > abc (a + b + c)

Dividing both sides by a²b²c², we have

⇒ \(\frac{1}{c^2}+\frac{1}{a^2}+\frac{1}{b^2}\) > \(\frac{a+b+c}{abc}\)

⇒ \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\) > \(\frac{2s}{abc}\)

⇒ \(\frac{2s}{abc}\) < \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\).