show this inequality with $a+b+c+d=1$ Let $a,b,c,d\ge 0$,and such $a+b+c+d=1$, show that
$$3(a^2+b^2+c^2+d^2)+64abcd\ge 1$$
use AM-GM 
$$a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$$
 it suffices to 
$$4\sqrt{abcd}+64abcd\ge 1$$
 A: Here is a way without Schur.  Let $p = a+b, u = ab$ and $q = c+d, v = cd$.  Then we have $p+q = 1$ and wish to show $$F = 3(p^2+q^2-2u-2v)+64uv \ge 1$$
For any fixed $p, q$, the variables $u, v$ can range continuously in $u \in [0, p^2/4]$ and $v \in [0, q^2/4]$.  Since $F$, viewed as a function of one variable (either $u$ or $v$) is linear, the minimum has to be when $u, v$ take one of the extreme points.  
Now if $u = p^2/4$ and $v = q^2/4$, the variables are equal and we have $F \ge 1 \iff (p-q)^2(p^2+q^2)\ge 0$ so we have equality when $p=q \implies a=b=c=d$ in this case.
Else we have $uv=0$, so WLOG let $d=0$ and the inequality is now the obvious
$$a+b+c=1 \implies 3(a^2+b^2+c^2) \ge 1$$
A: Here is a way using Schur and smoothing.  Let $a$ be the minimum among $a, b, c, d$ and say $3p = b+c+d, \,q = bc+cd+db, \,r = bcd$.  We define
$$F(a, b, c, d) = 3\sum_{cyc} a^2+64abcd $$
We will first show $F(a, b, c, d) \ge F(a, p, p, p)$.
$$\iff 3(b^2+c^2+d^2-3p^2) \ge 64a(p^3-r)$$
But $b^2+c^2+d^2-3p^2 = 2(3p^2-q)$, so we have to show
$$\iff 3(3p^2-q) \ge 32a(p^3-r) \tag{1}$$
By Schur inequality
$$\sum_{cyc} b^3+3bcd \ge \sum_{cyc} bc (b+c) \implies \frac43p(3p^2-q) \ge p^3-r$$
so it is enough for $(1)$ to show
$$3(3p^2-q) \ge 32a\cdot  \frac43p(3p^2-q) \iff (3p^2-q)(9-128ap) \ge 0 $$ 
Now $3p^2 \ge q$ is easily shown and $9-128ap = 9-128(1-3p)p > 0 $ as $p \ge \frac14$ for $a$ to be the minimum.  Hence we have shown $(1)$ holds true.
Now all that is needed is to show $F(1-3p, p, p, p) \ge 1$.
$$\iff 3((1-3p)^2+3p^2)+64(1-3p)p^3 \ge 1$$
$$\iff 2(1+2p)(1-3p)(1-4p)^2 \ge 0$$
which is now obvious. Equality is iff $3p=1$, i.e. $a=0, b=c=d=\frac13$ or when $4p=1$, i.e. $a=b=c=d=\frac14$.
A: the inequality equals:
$3(a+b+c+d)^2(a^2+b^2+c^2+d^2)+64abcd-(a+b+c+d)^4\ge 0 $ 
WLOG, let $d=min(a,b,c,d)$
$3(a+b+c+d)^2(a^2+b^2+c^2+d^2)+64abcd-(a+b+c+d)^4=2d\left(\dfrac{1}{3}\sum_{cyc (a,b,c)}(a-d)^2(2a+d)+\sum_{cyc (a,b,c)} a(a-d)^2+(a+b+c-3d)((a-b)^2+(b-c)^2+(a-c)^2)+3(a^3+b^3+c^3+3abc-\sum ab(a+b))+\dfrac{a^3+b^3+c^3}{3}-abc\right)+(a+b+c-3d)^2((a-b)^2+(b-c)^2+(a-c)^2)\ge0$
A: Maybe a generalization (not necessarily an easy way toward a solution):
Let $M_k=\sqrt[k]{(a^k+b^k+c^k+d^k)/4}$ be a $k$-th mean of the numbers, and $M_0=\sqrt[4]{abcd}$ the limit special case (geometric mean). Note that $M_1=1/4$. The question then is
$$3\cdot 4 M_2^2+64M_0^4\geq 1$$
$$\frac{3 M_2^2+M_0^4/M_1^2}{4}\geq M_1^2$$
$$\sqrt[4]{\frac{3 M_2^2M_1^2+M_0^4}{4}}\geq M_1$$
I recognized and the terms for the arithmetic mean so the expression is now scale-invariant (independent on the nurmalization of the original set of numbers).
What we got is a successive averaging of different means. Let's call $f_k$ the function that computes the $k$-th mean of its arguments. First compute means $M_0=f_0(a,b,\ldots)$, $M_1$ and $M_2$ (analogously). Then apply additional averaging in the following order:
$$f_4(f_4(f_0(M_1,M_2),M_0),f_0(M_1,M_2))\geq M_1$$
Whatever you make from this, I'm interested to know.
