2010/063) prove the inequality a+b/c+b+c/a+c+a/b>6 for a,b,c > 0 .?

this can be proved in at least 2 ways


method 1
we need to prove

(a+b)/c+(b+c)/a+(c+a)/b>6

or (a+b+c)/c + (a+b+c)/a + (a+b+ c)/b > 6

or (a+b+c) ( 1/a + 1/b+ 1/c) > 9

we know by AM GM inequality (a+b + c)/ 3 >= (abc)^(1/3)

again (1/a+ 1/b+ 1/c)/3 > = (1/abc)^(1/3)

multiplying we get (a+b + c)/ 3 * (1/a+ 1/b+ 1/c)/3 > = 1

or (a+b + c) * (1/a+ 1/b+ 1/c) >= 9

it is equal when a = b = c

proved


method 2
x+ 1/x >= 2 as (x+1/x) = (sqrt(x) -1/sqrt(x))^2 + 2

puttiing x = a/b . b/c and c/a in 3 times we get 3 eqautions

a/b + b/a >= 2
a/c+ c/a > = 2
b/c + c/b > = 2
adding above 3 and rearranging the terms we get the result.