Question

If A is the area of a triangle with side lengths a, b, and c, then​

Answer 1

Answer:

Step-by-step explanation:

If △ is the area of triangle with side lengths a,b,c, then show that △≤14⋅(a+b+c)abc−−−−−−−−−−−√. Also show that equality occurs in the above inequality when a=b=c

My attempt is as follows:-

△=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√

△=14(a+b+c)(b+c−a)(a+c−b)(a+b−c)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

b+c−a>0,a+c−b>0,a+b−c>0

So we can apply A.M≥G.M

b+c−a+a+c−b+a+b−c3≥((b+c−a)(a+c−b)(a+b−c))13

a+b+c3≥((b+c−a)(a+c−b)(a+b−c))13(1)

As a>0,b>0,c>0, we can also say

a+b+c3≥(abc)13(2)

But we can't say from this that (abc)13≥((b+c−a)(a+c−b)(a+b−c))13

So I tried G.M≥H.M for a,b,c

(abc)13≥3abcab+bc+ca

ab+bc+ca3≥(abc)23(3)

Seems like it didn't produce anything useful, so I tried G.M≥H.M for b+c−a,a+c−b,a+b−c

((b+c−a)(a+c−b)(a+b−c))13≥3(b+c−a)(a+c−b)(a+b−c)((b+c−a)(a+c−b)+(a+c−b)(a+b−c)+(b+c−a)(a+b−c))

((b+c−a)(a+c−b)(a+b−c))13≥3(b+c−a)(a+c−b)(a+b−c)2ab+2bc+2ca−a2−b2−c2(4)

But again it didn't produce anything useful, so I tried something else.

Applying A.M≥G.M for a+b+c,a+b−c,b+c−a,a+c−b

2(a+b+c)4≥((a+b+c)(b+c−a)(a+c−b)(a+b−c))14

a+b+c2≥((a+b+c)(b+c−a)(a+c−b)(a+b−c))14

Squaring both sides

(a+b+c)24≥((a+b+c)(b+c−a)(a+c−b)(a+b−c))12

By cachy's inequality:-

(12+12+12)(a2+b2+c2)≥(a+b+c)2

3(a2+b2+c2)≥(a+b+c)2

3(a2+b2+c2)4≥(a+b+c)24(5)

Hence

3(a2+b2+c2)4≥((a+b+c)(b+c−a)(a+c−b)(a+b−c))12(6)

Dividing by 4

3(a2+b2+c2)16≥△