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How does one prove that (a/sqrt(a 2 +8bc)) + (b/sqrt(b 2 +8ac)) + (c/sqrt(c 2 +8ab)) >= 1 ? sqrt() is the square root function

```
11 years ago 147 Points
```							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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```
11 years ago
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