<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] 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 $d=min(a,b,c,d)$
$3(a+b+c+d)^2(a^2+b^2+c^2+d^2)+64abcd-(a+b+c+d)^4=2d\left(\dfrac{1}{3}\sum_{cyc (a,b,c)}(a-d)^2(2a+d)+\sum_{cyc (a,b,c)} a(a-d)^2+(a+b+c-3d)((a-b)^2+(b-c)^2+(a-c)^2)+3(a^3+b^3+c^3+3abc-\sum ab(a+b))+\dfrac{a^3+b^3+c^3}{3}-abc\right)+(a+b+c-3d)^2((a-b)^2+(b-c)^2+(a-c)^2)\ge0$
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