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# How does one prove that(a/sqrt(a2+8bc)) + (b/sqrt(b2+8ac)) + (c/sqrt(c2+8ab)) >= 1 ?sqrt() is the square root function

## 1 Answers Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear Mardava RJgpl

((a/sqrt(a2+8bc)) + (b/sqrt(b2+8ac)) + (c/sqrt(c2+8ab)) >= 1

USe AM >=GM

((a/sqrt(a2+8bc)) + (b/sqrt(b2+8ac)) + (c/sqrt(c2+8ab))/3 >=((a/sqrt(a2+8bc)) (b/sqrt(b2+8ac))  (c/sqrt(c2+8ab))1/3    ............................1

Now use again AM>= GM

Let three number  a2,4bc,4bc

(a2 +4bc +4bc)/3 >=(16a2b2c2)1/3

(a2 +8bc ) >=3(16a2b2c2)1/3

or sqrt(a2 +8bc )>=√3(4abc)1/3

use this result in equation 1

((a/sqrt(a2+8bc)) + (b/sqrt(b2+8ac)) + (c/sqrt(c2+8ab)) >=3{abc/(33/24abc)}1/3

or

((a/sqrt(a2+8bc)) + (b/sqrt(b2+8ac)) + (c/sqrt(c2+8ab)) >=√3/41/3

>=1.09

>1

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