In my proof, I will use notation $(a,b,c,d)=(a_1,a_2,a_3,a_4)$ just for easier writing. Proof: Inserting $(a,b,c,d)=(1,1,4,9)$, we get: $K\leq \frac{11}{7}$. Let's prove inequality for $K=\frac{11}{7}$. WLOG we can suppose $a\le b\le c\le d$. Conditions are equivalent to: $d\ge (\sqrt{b}+\sqrt{c})^2$ and $a\leq \min\{(\sqrt{b}-\sqrt{c})^2,b\}$ Let's denote: $F(a,b,c,d)=\sum a^2 -\frac{11}{7}\sum_{sym} ab$ and let's fix $b$ and $c$ It's easy to see that: $\frac{\partial F}{\partial d}=2d-\frac{11}{7}(a+b+c)>0$ and $\frac{\partial F}{\partial a}=2a-\frac{11}{7}(b+c+d)<0$ From this, we conclude it's enough to prove inequality in case when $d= (\sqrt{b}+\sqrt{c})^2$ and $a=\min\{(\sqrt{b}-\sqrt{c})^2,b\}$ 1.st case: $b\leq c\leq 4b \Rightarrow a=(\sqrt{b}-\sqrt{c})^2$. Inequality is equivalent to: $(4b-c)(4c-b)\ge 0$ what is clearly true. 2.nd case: $c\ge 4b \Rightarrow a=b$. Inequality is equivalent to: $(\sqrt{c}-2\sqrt{b})(\sqrt{c}+3\sqrt{b})(3c+3\sqrt{bc}+2b) \ge 0$ and it's also true. $\blacksquare$

any elementary solution,please help me !

If we put $a_1=9, a_2=4, a_3=a_4=1$, then it's clear that $K \leq \frac{11}{7}$. Let's prove that inequality for $K=\frac{11}{7}$. Multiplying on 22, we see that we need to prove that \[11(a_1^2+a_2^2+a_3^2+a_4^2) \geq 22\sum_{i < j}a_ia_j.\] The main idea of this proof is to get this inequality as a sum of more simple inequalities. Let's notice that for $a\geq b \geq c \geq 0$ the inequality $a^2+b^2+c^2 >= 2(ab+bc+ca)$ is equivalent to $\sqrt{a} \geq \sqrt{b} + \sqrt{c}$. So we also get that $a_1\geq 9a_4$ and $a_2 \geq 4a_4$. So the list of "inequalities" we sum is: 1) $11(a_1^2+a_2^2+a_3^2) \geq 22(a_1a_2+a_2a_3+a_1a_3)$, 2) $\frac{14}{81}a_1^2+14a_4^2 \geq \frac{28}{9}a_1a_4$, 3) $3a_2^2=3a_2(\sqrt{a_2})^2 \geq 3a_2(\sqrt{a_3} + \sqrt{a_4})^2 \geq 12a_2a_4$, 4) $3a_3^2\geq3a_3a_4$, 5) $\frac{170}{81}a_1^2\geq\frac{170}{9}a_1a_4$, 6) $\frac{19}{81}a_1^2=\frac{19}{81}(\sqrt{a_1})^4\geq\frac{19}{81}(\sqrt{a_2}+\sqrt{a_3})^4\geq \frac{19}{81}(2\sqrt{a_3}+\sqrt{a_4})^4\geq\frac{19}{81}\left(\frac32\left(\sqrt{a_3}+\sqrt{a_4}\right)\right)^4 \geq 19a_3a_4$, 7) $\frac{40}{81}a_1^2=\frac{40}{81}(\sqrt{a_1})^4\geq\frac{40}{81}(\sqrt{a_2}+\sqrt{a_3})^4\geq \frac{40}{81}(\sqrt{a_2}+\sqrt{a_4})^4= \frac{40}{81}a_2a_4\left(\left(\frac{a_2}{a_4}\right)^{\frac14}+\left(\frac{a_4}{a_2}\right)^{\frac14}\right)^4\geq \frac{40}{81}a_2a_4\left(\left(4\right)^{\frac14}+\left(\frac14\right)^{\frac14}\right)^4 = \frac{40}{81}a_2a_4\left(\frac{3\sqrt{2}}{2}\right)^4=10a_2a_4.$