If $f$ is convex, is $f^{-1}$ concave? 
If $f:[0,+\infty)\to [0,+\infty)$ is a convex bijective function, is $f^{-1}$ a concave function? Or under what conditions $f^{-1}$ is concave?

Edit: 
What about a general bijective function $f:[a,b]\to[c,d]$ under any condition we will have $f$ convex then $f^{-1}$ concave 
Thank you very much 
 A: People have said that if $f$ is convex and increasing then yes. If $f$ is convex and a bijection (from $[0,\infty)$ to itself) then it is increasing:
Since $f$ is continuous on $(0,\infty)$, if we had $f(a)=0$ and $a>0$ then $f$ could not be a bijection. So $f(0)=0$. Now convexity implies that the function $f(x)/x$ is increasing, hence $f$ is increasing.
A: Since $f$ is convex then it is continuous. Since $f$ is injective it must be either increasing or decreasing. If $f$ is increasing then the answer is yes. Notice that in that case $f^{-1}$ is also increasing.    
Proof: Say $$f(tx_1+(1-t)x_2)\leq tf(x_1)+(1-t)f(x_2)$$
Let $y_i = f(x_i)$, then
$$(f(tx_1+(1-t)x_2))\leq f^{-1}(tf(x_1)+(1-t)f(x_2))$$
so 
$$tx_1+(1-t)x_2\leq f^{-1}(ty_1+(1-t)y_2)$$
Now since $x_i = f^{-1}(y_i)$ we finnaly have:
$$tf^{-1}(y_1)+(1-t)f^{-1}(y_2)\leq f^{-1}(ty_1+(1-t)y_2)$$
and we are done.
A: It is not always that the inverse of a convex function is convex. For example, if $f(x)=\frac{1}{x}$ then $f^{-1}(x)=\frac{1}{x}$. Both are convex.
Indeed, if $f(x)$ is convex and increasing then $f^{-1}(x)$ is concave. Mathematically, let $a = f(x) \implies f^{-1}(a)=x$ and $b=f(y) \implies y=f^{-1}(b)$. Since
$$f(\lambda x+(1-\lambda)y) \le \lambda f(x)+(1-\lambda)f(y), \tag{1}$$
we have
$$\lambda f^{-1}(a)+(1-\lambda)f^{-1}(b) \le f^{-1}\left(\lambda a+(1-\lambda)b\right). \tag{2}$$
Here, I have used the fact that $f(x)$ is increasing.
Can you figure out what happens when $f(x)$ is convex and decreasing?
A: The following uses the $f''>0$ convexity test. Therefore it is not a proof "from scratch", but indicates how the signs go in an automatic way. If $g:\>y\mapsto x$ is the inverse of $f:\>x\mapsto y$ then $f\bigl(g(y)\bigr)\equiv y$ and therefore $f'\bigl(g(y)\bigr)\cdot g'(y)\equiv1$. It follows that
$$f''\bigl(g(y)\bigr)g'^2(y)+f'\bigl(g(y)\bigr)g''(y)\equiv0\ ,$$
hence
$$g''(y)=-{f''(x)\over \bigl(f'(x)\bigr)^3}\ .$$
