Solution:

We have,

So,

=

 

So, we need to prove,

<

< 2

< 2

2abc-{(a+b-c)(b+c-a)(c+a-b)}>0..(1)

To prove (1), we have,

a2a2-(b-c)2

(a-b+c)(a+b-c)

b2b2-(c-a)2

(b-c+a)(b+c-a)

c2c2-(a-b)2

(c-a+b)(c+a-b)

a2b2c2 (a-b+c)2(b-c+a)2(c-a+b)2

abc (a-b+c)(b+c-a)(c+a-b) [Inequality holds when a=b=c]

Hence, 2abc (a-b+c)(b-c+a)(c-a+b)

2abc-(a-b+c)(b-c+a)(c-a+b)>0

Proved.