prove that the following function is convex at least I want a hint Let we have the following function 
$$ψ:R^{*+} \to R$$
        $$x\to ψ(x)=x^3$$
 How can I prove that $ψ$ is a convex function by using the definition ?
I meant that  I have to prove that 
$$ψ(λx+(1-λ)y) \le λψ(x)+ (1-λ)ψ(y)$$ where $λ \in [0,1] $ $x$ and $y$ belong to $R^{*+}$

Take $a,b$ positive reals. Then for $a \ge b$, we have $a^2+ab \ge 2b^2$, that is, $a^2 + ab+b^2 \ge 3b^2$. Multiplying by $(a-b)$, we get $a^3 - b^3 \ge 3b^2(a-b)$.
For $a\le b$, we have $a^2+ab \le 2b^2$, that is, $a^2+ab+b^2 \le 3b^2$. Here $(a-b) \le 0$, so multiplying by $(a-b)$, we have $a^3-b^3 \ge 3b^2(a-b)$.
Thus for any positive reals $a,b$, we have $a^3-b^3 \ge 3b^2(a-b)$, that is, $a^3 \ge b^3 + 3b^2(a-b). \tag{1}$
Taking $a = x, b = \lambda x + (1- \lambda)y$ in the above inequality, we get
$$ x^3 \ge (\lambda x + (1-\lambda)y)^3 + 3b^2(1-\lambda)(x-y) \tag{2}$$
Taking $a = y, b = \lambda x + (1- \lambda)y$ in $(1)$, we get 
$$ y^3 \ge (\lambda x + (1-\lambda)y)^3 + 3b^2(-\lambda)(x-y) \tag{3}$$
Now, $\lambda \times (2) + (1-\lambda) \times (3)$ gives
$$\lambda x^3 + (1-\lambda)y^3 \ge (\lambda x + (1-\lambda)y)^3 $$
That proves convexity the way you want. 
Note though that everything we did above follows from Taylor's theorem and non-negativity of the second derivative and is in fact a proof in disguise of the fact that a non-negative second derivative implies convexity. You are better off just using that. 
