Find all functions f : IR ---> IR : ∀ (x,y) ∈ IRē :
If we replace $\displaystyle f(t) ~~$ by $\displaystyle \frac{1}{g(t)}$
Then it becomes
$\displaystyle \frac{1}{g(\frac{x+y}{2})}[ g(y) + g(x) ] = 2$
$\displaystyle g(x) + g(y) = 2 g( \frac{x+y}{2})$
It implies that $\displaystyle g(t) = t ~~$ so $\displaystyle f(t) = \frac{1}{t}$
Is there another function satisifying the requirement ?
The equation $\displaystyle g\left(\frac{x+y}2\right)=\frac{g(x)+g(y)}2$ that occurs in simplependulum's reply is known as Jensen's equation.
To solve it, put $\displaystyle y=0$ and then $\displaystyle g\left(\frac x2\right)=\frac{g(x)+b}2$ where $\displaystyle b=g(0)$.
Put $\displaystyle x+y$ in this latter equation: $\displaystyle g\left(\frac{x+y}2\right)=\frac{g(x+y)+b}2$.
And now we see that $\displaystyle g(x+y)+b=g(x)+g(y)$.
Finally, put $\displaystyle h(x)=g(x)-b$. Then $\displaystyle h(x+y)=h(x)+h(y)$. This is the well-known Cauchy's equation.
The most general continuous solution of Cauchy's equation is $\displaystyle h(x)=ax$ where $\displaystyle a$ is constant. Therefore $\displaystyle g(x)=ax+b$ is the most general continuous solution of Jensen's equation.
However, it must be said that there exist solutions of Cauchy's equation on $\displaystyle \mathbb R$ which are not continuous, and they turn out to be extremely pathological. For example, if $\displaystyle h$ is one of these solutions and $\displaystyle I=(a,b)$ is any open interval then $\displaystyle h(I)$ is dense in $\displaystyle \mathbb R$.
So it makes sense in many cases to ask for conditions on the solution, such as continuity, or boundedness on a finite interval, or monotonicity, anything which would avoid these weird functions.