This should be enough for the ones who want to continue.

CatalinBordea wrote: Determine the largest value the expression $$ \sum_{1\le i<j\le 4} \left( x_i+x_j \right)\sqrt{x_ix_j} $$may achieve, as $ x_1,x_2,x_3,x_4 $ run through the non-negative real numbers, and add up to $ 1. $ Find also the specific values of this numbers that make the above sum achieve the asked maximum. From $\left ( \sqrt{x_i}-\sqrt{x_j} \right )^4 \ge 0$ we get $x_i^2+6x_ix_j+x_j^2 \ge 4\left( x_i+x_j \right)\sqrt{x_ix_j}$ Therefore $$ \sum_{1\le i<j\le 4} \left( x_i+x_j \right)\sqrt{x_ix_j} \le \frac{1}{4} \sum_{1\le i<j\le 4} \left( x_i^2+6x_ix_j+x_j^2\right)=\frac{3\left( x_1+x_2+x_3+x_4 \right)^2}{4}=\frac{3}{4}$$

I think the equality case holds for $x_{1}=x_{2}=x_{3}=x_{4}=\frac{1}{4}$.

sqing wrote:

Second line sum sembols shouldn't be cyclic.

Generalization 1 Let $x_{1},x_{2},x_{3},x_{4}$ be positive reals with sum $\lambda $($n,k\in \mathbf{R^+}$). Then prove that $$\sum_{1\leq i<j\leq 4}{\left(x_{i}+x_{j}\right)^{p}\sqrt[k]{x_{i}x_{j}}}\geq \dfrac{\sqrt[k]{9\lambda ^{4}}}{2^{2p}.\sqrt[2k]{\left(\sum_{1\leq i<j\leq 4}{x_{i}x_{j}}\right)^{kp-2}}}$$

Generalization 2 Let ($n,k\in \mathbf{R^+}$) $x_{1},x_{2},\cdots,x_{n}$ be positive reals with sum $\lambda $. Then prove that $$\sum_{1\leq i<j\leq n}{\left(x_{i}+x_{j}\right)^{p}\sqrt[k]{x_{i}x_{j}}}\leq \dfrac{(n-1)^p.\lambda^{2p}}{2^p\sqrt{\left(2n-4\right)^{p}}.\sqrt[2k]{\left(\sum_{1\leq i<j\leq 4}{x_{i}x_{j}}\right)^{kp-2}}}$$