For $x\ge0$ we have \begin{align}\frac x{1+x^2}\le\frac{12x+4}{25}\iff(2x-1)^2(3x+4)\ge0,\tag1\end{align}which is true. So \[E\le\sum_{k=1}^n\frac{12a_k+4}{25}=\frac{12\times\frac n2+4n}{25}=\frac{2n}5.\]The greatest value is achieved when $(1)$ reaches equality, or \[a_1=a_2=\cdots=a_n=\frac12,\]which indeed works!

This is the tangent line method?

@above yes

Different proof. First, by AM-GM, we have \[ a_i^2 + 1 = a_i^2 + \frac14+\cdots+\frac14 \ge 5\cdot 2^{-\frac{8}{5}}a_i^{\frac{2}{5}}, \]so that \[ E \le \frac15\cdot 2^{\frac{8}{5}}\sum_{1\le i\le n}f(a_i),\quad\text{where}\quad f(a_i)=a_i^{\frac{2}{5}}. \]Clearly $f$ is concave. So by Jensen's, \[ \sum_{1\le i\le n}f(a_i) \le n \cdot f\left(\frac1n\sum_{1\le i\le n}a_i\right) = nf(1/2) = n\cdot 2^{-\frac35}. \]Combining the last two displays, we immediately get \[ E\le \frac15\cdot 2^{\frac{8}{5}}\cdot n\cdot 2^{-\frac35} = \frac{2n}{5}, \]with equality iff $a_i=1/2$ for all $i$.

$E= \sum \frac{a_i}{1+a_i^2} \leq \sum \frac{a_i}{\frac{3}{4}+a_i}=n-\frac{3}{4} \sum \frac{1}{\frac{3}{4}+a_i} \leq n -\frac{3}{4} *\frac{n^2}{\frac{3}{4}n+\sum a_i}= \frac{2n}{5}$

https://artofproblemsolving.com/community/c6h1389313p7741977

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augustin_p wrote: Let $ n $ $(n\geq2)$ be an integer. Find the greatest possible value of the expression $$E=\frac{a_1}{1+a_1^2}+\frac{a_2}{1+a_2^2}+\ldots+\frac{a_n}{1+a_n^2}$$if the positive real numbers $a_1,a_2,\ldots,a_n$ satisfy $a_1+a_2+\ldots+a_n=\frac{n}{2}.$ What are the values of $a_1,a_2,\ldots,a_n$ when the greatest value is achieved? Sorry for the non -latex form. Let a_1 be the minimum of those n terms. Then E is smaller than F=n*a_1/((a_1)²+1). Also, since the sum is n/2, we get that a_1≤1/2. Note that taking the derivative of x/(x²+1) where x<1 ,we deduce that it increases as x increases. Thus F takes its maximal value whenever a_1=1/2. So we have the equality case in all equations==>all terms are equal to each other. Therefore E(Max)=2n/5 where a_1=a_2=.........=a_n=1/2.