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.
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Answered by
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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..
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