Prove a convex function

Question

I have to prove that if $f:A \to \mathbb{R}$ is convex and $c \ge 0$ then $c \cdot f:A \to \mathbb{R}$ is convex.

I know that function $f:A \to \mathbb{R}$ is convex if for $\forall x,y \in A$ and $\forall \lambda \in [0,1]:$ $f(\lambda x +(1- \lambda)y) \le \lambda f(x)+ (1-\lambda)f(y)$, but I don't know how to use this in this prove.

Is it ok to say:

$f(\lambda x +(1- \lambda)y) \le c \cdot f(\lambda x +(1- \lambda)y)$

$\lambda f(x)+ (1-\lambda)f(y) \le \lambda c \cdot f(x)+ (1-\lambda) c \cdot f(y)$

there for:

$c \cdot f(\lambda x +(1- \lambda)y) \le \lambda c \cdot f(x)+ (1-\lambda) c \cdot f(y)$

Answer 1

Since $c\geq 0$ and $f(\lambda x +(1- \lambda)y) \le \lambda f(x)+ (1-\lambda)f(y)$, multiplying by $c$ we obtain: $$ cf(\lambda x +(1- \lambda)y) \le \lambda cf(x)+ (1-\lambda)cf(y).$$ This is exactly the definition of $c.f$ is convex.