Let `agt0, bgt0, cgt0` and `a+b+c=6` then `((ab+1)^(2))/(b^(2))+((bc+1)^(2))/(c^(2))+((ca+1)^(2))/(a^(2))` may be

Options

A. `75/4`
B. `35`
C. `15`
D. `10`

Answer

A, B

Explanation

A.M. `ge` H.M for `1/a, 1/b, 1/c` we get `(1/a+1/b+1/c)/3 ge 3/6`
`implies 1/a+1/b+1/c ge 3/2`
Now, `((a+1/b)^(2)+(b+1/c)^(2)+(c+1/a)^(2))/3 ge ((a+1/b+b+1/c+c+1/a)/3)^(2) ge ((6+3/2)/3)^(2)`
`(a+1/b)^(2)+(b+1/c)^(2)+(c+1/a)^(2) ge 75/4`