<s>[INST] 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 
$$4\sqrt{abcd}+64abcd\ge 1$$
 [/INST] 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
$$a+b+c=1 \implies 3(a^2+b^2+c^2) \ge 1$$
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