Let $a,b,c,d>0$. Prove that$:$ $$3(a^4 + b^4 + c^4 + d^4) + 4abcd \geqslant (\,a+b+c+d\,)(\,a^3 + b^3 + c^3 + d^{3}\,)$$
As pointed out by @tthnew, this question was posted in:
https://artofproblemsolving.com/community/c6h2233061p17081874
There are some solutions there. Also, @Michael Rozenberg said "There is a nice solution by SOS and without computer of course."
As pointed out by @Maximilian Janisch (https://rgmia.org/papers/v8n2/M4.pdf), it is Suranyi's inequality for $n=4$:
(Janos Suranyi) If $x_k > 0$ ($k=1, 2, \cdots, n$), then the following inequality holds:
$$(n-1)\sum_{k=1}^n x_k^n + n\prod_{k=1}^n x_k \ge \sum_{k=1}^n x_k \sum_{k=1}^n x_k^{n-1}.$$
I tried to use the Turcevici's inequality:
$$a^4+b^4+c^4+d^4+2abcd\geq a^2b^2+a^2c^2+b^2c^2+a^2d^2+b^2d^2+c^2d^2$$
It remains to prove
$$a^2b^2+a^2c^2+b^2c^2+a^2d^2+b^2d^2+c^2d^2−2abcd+2abcd\geq\frac{1}{2}\sum_{cyc}a^3(b+c+d),$$
but it's wrong(the reversed is true). Help me please.
$$3(a^4 + b^4 + c^4 + d^4) + 4abcd-(a+b+c+d)(a^3 + b^3 + c^3 + d^3)=$$ $$=\sum_{cyc}(2a^4+abcd-a^3(b+c+d))=\frac{1}{6}\sum_{sym}(2a^4+abcd-3a^3b)=$$ $$=\frac{1}{6}\sum_{sym}(2a^4-a^3b-ab^3-ab^3+abcd)=\frac{1}{6}\sum_{sym}((a-b)^2(a^2+ab+b^2)-a^3b+abcd)=$$ $$=\frac{1}{12}\sum_{sym}((a-b)^2(2a^2+2b^2+2ab)-a^3b-ab^3+2a^2b^2-2a^2b^2+2a^2bc-2a^2bc+2abcd)=$$ $$=\frac{1}{12}\sum_{sym}((a-b)^2(2a^2+2b^2+ab)-d^2(a-b)^2-cd(a^2-ab)-cd(b^2-ab))=$$ $$=\frac{1}{24}\sum_{sym}(2(a-b)^2(2a^2+2b^2+ab)-c^2(a-b)^2-d^2(a-b)^2-2cd(a-b)^2)=$$ $$=\frac{1}{24}\sum_{sym}(a-b)^2(4a^2+4b^2+2ab-(c+d)^2).$$ Now, let $a\geq b\geq c\geq d$, $S_{ab}=4a^2+4b^2+2ab-(c+d)^2,$...
Also, easy to see that $$S_{ad}=4(a^2+d^2)+2ad-(b+c)^2\geq0,$$ $$S_{ac}=4(a^2+c^2)+2ac-(b+d)^2\geq0,$$ $$S_{ab}=4(a^2+b^2)+2ab-(c+d)^2\geq0,$$ $$S_{ac}+S_{bc}=4(a^2+c^2)+2ac-(b+d)^2+4(b^2+c^2)+2bc-(a+d)^2=$$ $$=4(a^2+c^2)+2ac-(a+d)^2+4(b^2+c^2)+2bc-(b+d)^2\geq0$$ and $$b^2((a-d)^2S_{ad}+(b-d)^2S_{bd})\geq a^2(b-d)^2S_{ad}+b^2(b-d)^2S_{bd}=$$ $$=(b-d)^2(a^2(4a^2+2ad+4d^2-(b+c)^2+b^2(4b^2+2bd+4d^2)-(a+c)^2))\geq$$ $$\geq(b-d)^2(4(a^2-b^2)^2+6a^2b^2-2a^2bc-2b^2ac-a^2c^2-b^2c^2)\geq$$ $$\geq(c-d)^2(4(a^2-b^2)^2+6a^2b^2-2a^2bc-2b^2ac-a^2c^2-b^2c^2).$$
Thus, since $$\frac{\partial}{\partial a}(4a^4-3a^2b^2-2ab^3+3b^4-2a^2bc-2b^2ac-a^2c^2+3b^2c^2)=$$ $$=16a^3-6ab^2-2b^3-4abc-2b^2c-2ac^2\geq0,$$ we obtain: $$b^2\left((a-b)^2S_{ab}+(a-c)^2S_{ac}+(a-d)^2S_{ad}+(b-c)^2S_{bc}+(b-d)^2S_{bd}+(c-d)^2S_{cd}\right)\geq$$ $$\geq(c-d)^2\left(4(a^2-b^2)^2+6a^2b^2-2a^2bc-2b^2ac-a^2c^2-b^2c^2+b^2S_{cd}\right)\geq$$ $$\geq(c-d)^2\left(4(a^2-b^2)^2+6a^2b^2-2a^2bc-2b^2ac-a^2c^2-b^2c^2+4b^2c^2-b^2(a+b)^2\right)=$$ $$=(c-d)^2(4a^4-3a^2b^2-2ab^3+3b^4-2a^2bc-2b^2ac-a^2c^2+3b^2c^2)\geq$$ $$\geq(c-d)^2(4b^4-3b^4-2b^4+3b^4-2b^3c-2b^3c-b^2c^2+3b^2c^2)=$$ $$=2b^2(b-c)^2(c-d)^2\geq0$$ and we are done!