Turkevicius inequality

Assume $a\geq b\geq c\geq d.$ Let $f(a) = a^4+b^4+c^4+d^4+2abcd - a^2b^2-a^2c^2-a^2d^2-b^2c^2-b^2d^2-c^2d^2.$ Then, $f'(a) = 4a^3 -2a(b^2+c^2+d^2)+2bcd$ and $f''(a) = 12a^2 - 2b^2-2c^2-2d^2\geq 0.$ Hence, $f'$ is increasing and $f'(b) = 2(b^3-b(c^2+d^2)+bcd) \geq 2b(b-c)(b-d)\geq 0.$ Thus, $f(a)$ is increasing and therefore:

$$f(a)\geq f(b) = b^4+c^4+d^4+2b^2cd-2b^2c^2-2b^2d^2-c^2d^2 = $$$$b^4-2b^2(c^2+d^2-cd)+c^4+d^4-c^2d^2 = (b^2 - c^2-d^2+cd)^2+2cd(c-d)^2\geq 0.$$

Let $a\geq b\geq c\geq d$.

Thus, $$d(d-a)(d-b)(d+2c)+d(d-a)(d+2b)(d-c)+d(d+2a)(d-b)(d-c)\geq0$$ or $$d^4-(ab+ac+bc)d^2+2abcd\geq0.$$ Thus, it's enough to prove that $$a^4+b^4+c^4\geq(a^2+b^2+c^2-ab-ac-bc)d^2+a^2b^2+a^2c^2+b^2c^2$$ or $$\sum_{cyc}(a^2-b^2)^2\geq\sum_{cyc}(a-b)^2d^2$$ or $$\sum_{cyc}(a-b)^2((a+b)^2-d^2)\geq0,$$ which is obvious.

Here $\sum\limits_{cyc}$ means $\sum\limits_{a\rightarrow b\rightarrow c\rightarrow a}$.

Also, we can use the BW.

Let $a=\min\{a,b,c,d\}$, $b=a+u$, $c=a+v$ and $d=a+w$.

Thus, $u$, $b$ and $w$ are non-negatives and $$\sum_{cyc}a^4+2abcd-\frac{1}{6}\sum_{sym}a^2b^2=$$ $$\sum_{cyc}(3u^2-2uv)a^2+\sum_{cyc}\left(4u^3-u^2v-u^2w+\frac{2}{3}uvw\right)a+\sum_{cyc}(u^4-u^2v^2)\geq0.$$

Also there are proofs by EV Method and by SOS.