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 
$$4\sqrt{abcd}+64abcd\ge 1$$

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. $a=b=c=d=\frac14$.
