Find the maximum value of the expression ${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$. 
Find the maximum value of the expression :
  $${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$$
  where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$.

Taking $x=y=z=\frac {1}{3}$, I get the expression as $\frac {3x}{1+x^2}$, which is equal to $\frac {1}{1+{\frac{1}{9}}}$ or $\frac {9}{10}$.
How can I actually solve the problem without making unnecessary assumptions ?

We use the method of Lagrange Multipliers.
First, we define the function to maximize as $f(x,y,z)=\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ and our constraint as $g(x,y,z)=x+y+z-1$.
Thus, the Lagrange function is:
$$\mathcal{L}(x,y,z,\lambda)=f(x,y,z)+\lambda\cdot g(x,y,z)$$
$$\mathcal{L}(x,y,z,\lambda)=\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}+\lambda(x+y+z-1) \tag{1}$$
Where $\lambda$ is some scalar. This function's partial derivatives must equal all equal to zero:
$$\frac{\partial \mathcal{L}}{\partial x}=\frac{1-x^2}{(x^2+1)^2}+\lambda=0$$
$$\frac{\partial \mathcal{L}}{\partial y}=\frac{1-y^2}{(y^2+1)^2}+\lambda=0$$
$$\frac{\partial \mathcal{L}}{\partial z}=\frac{1-z^2}{(z^2+1)^2}+\lambda=0$$
$$\frac{\partial \mathcal{L}}{\partial \lambda}=x+y+z-1=0$$
Solving this system of equations will yield values for $x,y,z,\lambda$ which will either minimize or maximize $f(x,y,z)$.
After some tedious calculation, we notice that the solution which maximizes the value of $f(x,y,z)$ is:
$x=y=z=\frac{1}{3}$ and $\lambda=-0.72$, as you have correctly predicted.
Thus, the maximum value of $f(x,y,z)=\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}=\frac{9}{10}$.
