if D2f(x)>0 and Df(x) + f(x) >0 prove that f(x) >0?

Question

D2f(x) is the second derivative of f(x)

Answer 1

Let f(x) = x^2-1 and x = 1.
Df(x) = 2x
D2f(x) = 2 > 0. as required
Df(1) + f(1) = 2 + 0 = 2 > 0. also as required.
f(1) = 0 which is not greater than 0.

Your statement is incorrect by counterexample.

Answer 2

This is not right.
Counter example: f(x) = x^2 - 10 at x = 3
f'(3) + f(3) = 2*3 + 3^2 - 10 = 5 > 0
f"(3) = 2 > 0
But f(3) = 3^2 - 10 = -1 < 0