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 $x + y + z = 1 $.
My Attempt
$$
\begin{align}
\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}&amp;=\frac{1}{x+\frac{1}{x}}+\frac{1}{y+\frac{1}{y}}+\frac{1}{z+\frac{1}{z}}\\&amp;\geq \frac{(1+1+1)^2}{x+\frac{1}{x}+y+\frac{1}{y} + z+\frac{1}{z}}\\&amp;=\frac{9}{x+y+z +(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}
\end{align}
$$
From AM-HM 
$$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 3^2 $$
$$\implies (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 9 $$
$$\implies (x+y+z)+ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 10$$
From this point i am not being able o proceed as inequality signs are getting mixed. Can anyone help out? Thanks in advance.
You've already been given a hint to try Jensens inequality, which works fine in this case. Alternatively by AM-GM
$$9\frac{x}{9+(3x)^2}\leqslant 9\frac{x}{10(3x)^{2/10}}=\frac3{10}(3x)^{4/5}$$
Sum that up and observe by  power means:
$$\sum (3x)^{4/5} \leqslant 3\left(\frac{3x+3y+3z}3\right)^{4/5}=3$$
Equality is possible when $x=y=z=\frac13$, so these inequalities give you the maximum.
Ok rugi I will solve it using derivatives for your question consider x=y=z because for any condition like x+y+z=a we need to consider x=y=z to get maxima .So answer is 9÷10=0.9.This is for a &lt;3.And when the degree in the numerator is less than the degree in denominator in which the value is to be found out