Let $a,b,c,d\geq0$ satisfying $a+b+c+d=4$ . Prove $$\sqrt{a^3+b^3+c^3+d^3}+2(\sqrt3 -1)abcd\geq\sqrt{3(abc+abd+acd+bcd)}$$
Attempt: $a^3+b^3+c^3+d^3=(a+b+c+d)(a^2+b^2+c^2+d^2-ab-bc-cd-da-ac-bd)+3(abc+abd+acd+bcd)$ .
Let's try it with Vasc's EV. Who knows?
Yes, EV helps here!
We need to prove that: $$\sum_{cyc}a^3+8(2-\sqrt3)a^2b^2c^2d^2+4(\sqrt3-1)abcd\sqrt{\sum_{cyc}a^3}\geq3\sum_{cyc}abc.$$
Now, by the Vasc's EV Method we can prove that: $$8(2-\sqrt3)a^2b^2c^2d^2+4(\sqrt3-1)abcd\sqrt{\sum_{cyc}a^3}\geq\frac{42abcd(a+b+c+d)^2}{5(a^3+b^3+c^3+d^3)+(a+b+c+d)^3}$$ or $$2(2-\sqrt3)abcd+(\sqrt3-1)\sqrt{\sum_{cyc}a^3}\geq\frac{168}{5(a^3+b^3+c^3+d^3)+64}.$$ Indeed, let $a^3+b^3+c^3+d^3=const.$
Thus, by corollary 1.8 (b) here:
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
$abcd$ gets a minimal value for an equality case of three variables.
Let $b=c=a$ and $d=4-3a$, where $0\leq a\leq \frac{4}{3}.$
Thus, we need to prove that: $$2(2-\sqrt3)a^3(4-3a)+(\sqrt3-1)\sqrt{3a^3+(4-3a)^3}\geq\frac{168}{5(3a^3+(4-3a)^3)+64},$$ which is true by hand, but here WA is better:
https://www.wolframalpha.com/input/?i=min%7B2%282-sqrt3%29x%5E3%284-3x%29%2B%28sqrt3-1%29sqrt%283x%5E3%2B%284-3x%29%5E3%29-168%2F%285%283x%5E3%2B%284-3x%29%5E3%29%2B64%29%7D%2C0%3Cx%3C4%2F3
Id est, it's enough to prove that: $$a^3+b^3+c^3+d^3+\frac{42abcd(a+b+c+d)^2}{5(a^3+b^3+c^3+d^3)+(a+b+c+d)^3}\geq3(abc+abd+acd+bcd),$$ which is true by BW:
Let $a=\min\{a,b,c,d\}$, $b=a+u,$ $c=a+v$ and $d=a+w$.
Thus, we need to prove that: $$4\sum_{cyc}(3u^2-2uv)a^4+\sum_{cyc}\left(35u^3-19u^2v-19u^2w+\frac{22}{3}uvw\right)a^3+$$ $$+\sum_{cyc}(36u^4-13u^3v-13u^3w+22u^2v^2-31u^2vw)a^2+$$ $$+\sum_{cyc}(15u^5-u^4v-u^4w+12u^3v^2+12u^3w^2-23u^3vw-14u^2v^2w)a+$$+ $$+\frac{1}{3}(5(u^3+v^3+w^3)+(u+v+w)^3)(u^3+v^3+w^3-3uvw)\geq0,$$ which is true by Schur and Muirhead.