Symmetric Cournot equilibrium: suffciency without second order conditon Let $q_i \in Q = \mathbb R_+$ denote the quantity produced by firm $i \in \{1,2\}$. Further let $\pi_i(q_1,q_2)  = (1-q_1-q_2)q_i$ denote the profits of $i$. A Nash equilibrium $(q_1^*,q_2^*) \in Q^2$ satisfies
\begin{align}
&amp;\pi_1(q_1^*,q_2^*) \geq \pi_1(q_1,q_2^*) \quad \forall q_1 \in Q\\
&amp;\pi_2(q_1^*,q_2^*) \geq \pi_2(q_1^*,q_2) \quad \forall q_2 \in Q. 
\end{align}
We are considering symmetric equilibria of the form $q^* = q_1^* = q_2^*$ and therefore apply the symmetric opponents form approach.
Define $\pi(q,q^*) = \pi_1(q,q^*)$.
There exists a unique symmetric root to the first order condition $\pi_q(q^*,q^*) = 0$ given by $q^* = \frac{1}{3}$.

Claim The candidate $q = \frac{1}{3}$ is the unique symmetric maximizer of $\pi(q,q^*)$.

Problem: The candidate might be a minimum or saddle.

The idea: In economic settings equilibrium quantities are basically restricted by individual rationality, i.e. $\pi(q^*,q^*) = (1-2q^*)q^*$ implies $q^* \in [0,\frac{1}{2}]$. Since $\pi(\frac{1}{3},\frac{1}{3}) = \frac{1}{9} &gt; 0$, the claim follows. 

Edit I edit the question to further clarify the issue. Suppose I don't have any information about concavity of $\pi(q,q^*)$ w.r.t. $q$. 

A general argument:
We need to distinguish 4 cases. 


$q^*$ is a saddle and $\pi(q^*,q^*) &gt; 0$ and $\pi(\infty,q^*) = \infty$.
$q^*$ is a saddle and $\pi(q^*,q^*) &lt; 0$ and $\pi(\infty,q^*) = -\infty$. 
$q^*$ is a minimum and $\pi(q^*,q^*) &lt; 0$ and $\pi(\infty,q^*) = \infty$.
$q^*$ is a maximum and $\pi(q^*,q^*) &gt; 0$ and $\pi(\infty,q^*) = -\infty$.


Since case 4 is considered here $\frac{1}{3} = \arg\max_q\pi(q,q^*)$.
How can it be a minimum or saddle since for every $q_2$ the profit function is strictly concave? If anything, the implicit non-negativity constraints might be binding, for example, if $q_2\geq 1/2$. 

When the objective function is strictly concave, the first order conditions are necessary and sufficient, and the maximum is unique. 

Not sure if I'm answering your question, but hopefully it helps.