<s>[INST] Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove $a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$ 
Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove  $$a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$$

The inequality can be written in the condensed form
$$\sum\limits_{Sym}a\ge\sum\limits_{Sym}a^3+\sum\limits_{Sym}ab$$
I was told that this is a pretty inequality to prove, but I have been unable to do so. 
I've tried naive things like multiplying both sides by $a+b+c+d$, and rewriting $(a^2+b^2+c^2+d^2)^2$, but nothing panned out (and the computations were relatively time-consuming). I also tried looking for clever applications of Cauchy-Schwarz (which seems like the way to go) and AM-GM, but nothing sprung out at me.
 [/INST] Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$ and $abc+abd+acd+bcd=4w^3$.
Hence, $16u^2-12v^2=1$ and our inequality is equivalent to $3v^6-4uv^2w^3+w^6\geq0$.
By Roll's theorem there are $x>0$, $y>0$ and $z>0$, for which 
$x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
After this substitution we need to prove that
$\sum\limits_{cyc}(x^3y^3-x^3y^2z-x^3z^2y+x^2y^2z^2)\geq0$, which is Schur.
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