Question

For positive real numbers a, b, c > 0, prove that:
(a² + b²)² ≥ (a + b - c) (a + b + c) (b + c - a) (a + c - b).
Pls help.

Answer 1

Refer to the attachement ...

Answer 2

Answer:

Given a

2

+b

2

+c

2

=1

we know that, (a+b+c)

2

≥0

⇒a

2

+b

2

+c

2

+2(ab+bc+ca)≥0

⇒1+2(ab+bc+ca)≥0

⇒ab+bc+ca≥

2

−1

Also, (b−c)

2

+(c−a)

2

+(a−b)

2

≥0

⇒2(a

2

+b

2

+c

2

)−2(ab+bc+ca)≥0

⇒ab+bc+ca≤a

2

+b

2

+c

2

⇒ab+bc+ca≤1

Hence

2

−1

≤ab+bc+ca≤1

Therefore, ab+bc+ca is less than 1.

ANSWERED BY MIND..