Turkevicius inequality From Zdravko Cvetkovski's Inequalities — Theorems, Techniques and Selected:

Let $a$, $b$, $c$ and $d$ be positive real numbers. Prove that:
$$a^4+b^4+c^4+d^4+2abcd \geq a^2b^2+ a^2c^2+ a^2d^2+ b^2c^2+ b^2d^2+ c^2d^2.$$


Please suggest how to prove this inequality using basic methods. No logarithms.

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. 
