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If ∆ is the area of a triangle with side lengths a, b, c, then show that ∆ ≤ 1/4 √ (a + b + c) abc. Also show that the equality occurs in the above inequality if and only if a = b = c.

If ∆ is the area of a triangle with side lengths a, b, c, then show that ∆ ≤ 1/4 √ (a + b + c) abc. Also show that the equality occurs in the above inequality if and only if
a = b = c.

Grade:11

1 Answers

Jitender Pal
askIITians Faculty 365 Points
6 years ago
Hello Student,
Please find the answer to your question
We know, ∆ = √s (s – a) (s – b) (s – c)
= √s/8 (b + c – a) (c + a – b) (a + b – c)
Since sum of two sides is always greater than third side;
∴ b + c – a, c + a – b, a + b – c > 0
⇒ (s – a) (s – b) (s – c) > 0
Let s – a = x. s – b = y, s – c = z
Now, x + y = 2 s – a – b = c
Similarly, y + z = a
And z + x = b
Since AM ≥ GM
⇒ x +y/2 ≥ √xy ⇒ 2√xy ≤ a y + z/2 ≥ √yz ⇒ 2√yz ≤ b z + x/2 ≥ √xz ⇒ 2√xz ≤ c ∴ 8xyz ≤ abc
⇒ (s – a) (s – b) (s – c) ≤ 1/8 abc
⇒ s (s – a) (s – b) (s – c) ≤ sabc/8
⇒ ≤ 1/16 (a + b + c) abc ⇒ ∆ ≤ 1/4 √abc (a + B + c)
And equality holds when x = y = z ⇒ a = b = c

Thanks
Jitender Pal
askIITians Faculty

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