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.
 A: 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.
A: I got some hints from Crux Problem 3059, click here for more details=)
Based on the hints (which tries to relate the inequality with a constrained optimization problem), I worked out a proof as follows:
We have
\begin{align}
\frac{1}{2}(a+b+c+d)^2=&\frac{1}{2}(a^2+b^2+c^2+d^2)+ab+ac+ad+bc+bd+cd\\
=&\frac{1}{2}+ab+ac+ad+bc+bd+cd
\end{align}
by assumption. Thus, in order to prove the inequality, it is equivalent to prove
$$
a+b+c+d\geq a^3+b^3+c^3+d^3+\frac{1}{2}(a+b+c+d)^2-\frac{1}{2}.
$$
This can further be simplified as proving
$$
a^3+b^3+c^3+d^3+\frac{1}{2}(a+b+c+d-1)^2\leq1.~~~~(*)
$$
Now we try to maximize the LHS under the constraint, i.e.
\begin{align}
\max&~~f\triangleq a^3+b^3+c^3+d^3+\frac{1}{2}(a+b+c+d-1)^2\\
s.t.&~~a^2+b^2+c^2+d^2=1.
\end{align}
Now we try to use the Lagrangian multiplier method. Let
$$
L=f+\lambda(a^2+b^2+c^2+d^2-1).
$$
Take the derivative of $L$ regarding $a,b,c,d$ respectively and let the derivative equal to $0$ gives the following set of equations:
\begin{align}
L_a=&a+b+c+d+3a^2+2a\lambda=1\\
L_b=&a+b+c+d+3b^2+2b\lambda=1\\
L_c=&a+b+c+d+3c^2+2c\lambda=1\\
L_d=&a+b+c+d+3d^2+2d\lambda=1,
\end{align}
Notice that these four equations share exactly the same form. Thus, either
$$
3x^2+2\lambda x=0,~~x=a,b,c,d,~~~~(case1)
$$
or
$$
a=b=c=d.~~~~(case2)
$$
We analyse these two cases separately:
Case 1:
Under this case, we have
$$
a+b+c+d=1.
$$
Since
$$
a^2+b^2+c^2+d^2=1,
$$
the only possibility is one of four elements $a,b,c,d$ equal to $1$ and others are all $0$, which gives $\lambda=-3/2$ and $f=1$.
Case 2:
By the assumption
$$
a^2+b^2+c^2+d^2=1,
$$
it follows
$$
a=b=c=d=\frac{1}{2},
$$
which gives $\lambda=-7/4$ and $f=1$.
Above all we have the maximum of $f$ is $1$ and this proves $(*)$.
