if a >b> 0 show that √ab lies between a and b
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Given, a2 + b2 + c2 = 1
We know that,
(a + b + c)2 ≥ 0
⇒ a2 + b2 + c2 + 2(ab + bc + ca) ≥ 0
⇒ 1 + 2(ab + bc + ca) ≥ 0
⇒ ab + bc + ca ≥ 
Again, (b - c)2 + (c - a)2 + (a - b)2 ≥ 0
⇒ 2(a2 + b2 + c2) - 2(ab + bc + ca) ≥ 0
⇒ ab + bc + ca ≤ a2 + b2 + c2
⇒ ab + bc + ca ≤ 1
Hence, 
We know that,
(a + b + c)2 ≥ 0
⇒ a2 + b2 + c2 + 2(ab + bc + ca) ≥ 0
⇒ 1 + 2(ab + bc + ca) ≥ 0
⇒ ab + bc + ca ≥ 
Again, (b - c)2 + (c - a)2 + (a - b)2 ≥ 0
⇒ 2(a2 + b2 + c2) - 2(ab + bc + ca) ≥ 0
⇒ ab + bc + ca ≤ a2 + b2 + c2
⇒ ab + bc + ca ≤ 1
Hence, 
Parkar:
there you have written 2 is that a power??
thanx
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