Let `x_1, x_2, ,x_n` be positive real numbers and we define `S=x_1+x_2++x_ndot` Prove that `(1+x_1)(1+x_2)(1+x_n)lt=1+S+(S^2)/(2!)+(S^3)/(3!)++(S^n)/(n !)`

Explanation

We have to prove that
`(1 + x_(2)) (1 + x_(2))…..(1 + x_(n)) le 1 + S + (S^(2))/(2!) + (S^(3))/(3!) + ……+ (S^(n))/(n!)`
Product on left hand side suggests that we must consider G.M of
`(1 + x_(1)), (1 + x_(2)),….., (1 + x_(n))`. Also, `(1 + x_(1))(1 + x_(2)),....., (1 + x_(n))` are postive. Now,
`A.M le G.M`
`implies ((1 + x_(1)) + (1 + x_(2)) + .....+ (1 + x_(n)))/(n) ge [(1 + x_(1))(1 + x_(2))......(1 + x_(n))]^((1)/(n)`
or `[(1 + x_(1))(1 + x_(2))......(1 + x_(n))]^(1//2) le (n + (x_(1) + x_(2) + ........ + x_(n)))/(n)`
or `(1 + x_(1)) (1 + x_(2)).....(1 + x_(n)) le ((n + S)/(n))^(n)`
Now `(1 + (S)/(n))^(n) = 1 + S + ((1 - (1)/(n)))/(2!) S^(2) + (1 - (1)/(n))(1 - (2)/(n))/(3!) S^(3) + .....`
Now, `(1 - (1)/(n))/(2!) lt 1, ((1 - (1)/(n))(1 - (2)/(n)))/(3!) lt 1` and so on.
Hence, `(1 + x_(1)) (1 + x_(2)) ....... (1 + x_(n)) le 1 + S + (S^(2))/(2!) + (S^(3))/(3!) + .....`