How does one prove that

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

sqrt() is the square root function

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/(3^3/24abc)}^1/3

or  

((a/sqrt(a²+8bc)) + (b/sqrt(b²+8ac)) + (c/sqrt(c²+8ab)) >=√3/4^1/3

                                                                                                            >=1.09

                                                                                                             >1