how to solve this problem. pls help
Answers
Step-by-step explanation:
Without loss of generality we can assume that a≥b≥c
Let
(−a+b+c)(a−b+c)(a+b−c)=S
⇒S=(−a+b+c)[a−(b−c)][a+(b−c)]
⇒S=(−a+b+c)[a
2
−(b−c)
2
]
⇒S=(−a+b+c)[a
2
−b
2
−c
2
+2bc]
⇒S=−(a
3
+b
3
+c
3
)−2abc+b
2
c+bc
2
+ab
2
+a
2
b+ac
2
+a
2
c
⇒abc−S=(a
3
+b
3
+c
3
)+3abc−(b
2
c+bc
2
+ab
2
+a
2
b+ac
2
+a
2
c)
⇒abc−S=(a
3
−a
2
b)+(b
3
−b
2
c)+(c
3
−c
2
a)+(abc−bc
2
)+(abc−ab
2
)+(abc−a
2
c)
⇒abc−S=a
2
(a−b)+b
2
(b−c)+c
2
(c−a)+bc(a−c)+ab(c−b)+ac(b−a)
⇒abc−S=a(a−b)(a−c)+b(b−c)(b−a)+c(c−a)(c−b)
⇒abc−S=(a−b)[a(a−c)−b(b−c)]+c(c−a)(c−b)
⇒abc−S=(a−b)2[a
2
−b
2
+c(b−a)]+c(c−a)(c−b)
⇒abc−S=(a−b)
2
[a+b−c]+c(c−a)(c−b)
Now (c−a)≤0 and (c−b)≤0
⇒c(c−a)(c−b)≥0
and
(a−b)
2
(a+b−c)≥0
This shows
abc−S≥0
⇒S−abc≤0⇒(b+c−a)(c+a−b)(a+b−c)−abc≤0
Hence, the expression is non-positive.
So, (C)