If a, b,c are three positive real numbers , then find minimum value of `(a^(2)+1)/(b+c)+(b^(2)+1)/(c+a)+(c^(2)+1)/(a+b)`

Explanation

Using `A.M ge G.M`, we get
`a^(2) + 1 ge 2a`
`implies (a^(2) + 1)/(b + c) ge (2a)/(b + c)`
Similarly, `(b^(2) + 1)/(a + c) ge (2b)/(a + c)`
and `(c^(2) + 1)/(a + b) ge (2c)/(a + b)`
Adding (1),(2) and (3), we get
`(a^(2) + 1)/(b + c) + (b^(2) + 1)/(a + c) + (c^(2) + 1)/(a + b) ge (2a)/(b + c) + (2b)/(b + c) + (2c)/(a + b)`
Now, `(2a)/(b + c) + (2b)/(a + c) + (2c)/(a + b)`
`= ((2a)/(b + c) + 2) + ((2b)/(a + c) + 2) + ((2c)/(a + b) + 2) - 6`
`= 2((a + b + c)/(b + c) + (a + b+ c)/(a + c) + (a + b + c)/(a + b))/(3)`
`ge (3)/((b + c)/(a + b+ c) + (a + c)/(a + b+ c) + (a + b)/(a + b+ c))`
`implies (a + b + c)/(b + c) + (a + b + c)/(a + c) + (a + b + c)/(a + b) ge (9)/(2)`
`implies (2a)/(b + c) + (2b)/(a + c) + (2c)/(a + b) ge 9 - 6 = 3`
`implies (a^(2) + 1)/(b + c) + (b^(2) + 1)/(a + c) + (c^(2) + 1)/(a + b) ge 3`
Hence, minimum value is 3.