<s>[INST] For positve $a$, $b$, $c$, $d$ with $a+b+c+d\leq 1$, prove that $\sqrt[4]{(1-a^4)(1-b^4)(1-c^4)(1-d^4)}\geq255\cdot a b c d .$ 
Let $a,b,c,d\in\mathbb R_+$ such that $a+b+c+d\leqslant1$. Prove that$$
\sqrt[4]{\smash[b]{(1-a^4)(1-b^4)(1-c^4)(1-d^4)}}\geqslant255·abcd.
$$

My observations:
I can see that all of $a,b,c,d$ are positive fractions which makes all the bracketed factors in LHS positive. Now if we raise both sides to the power $4$, and replace the $4$th powers of $a,b,c,d$ with $A,B,C,D$, our inequality gets reduced to
$$(1-A)(1-B)(1-C)(1-D)\ge 255^4 \cdot ABCD$$
Now applying AM $\ge$ GM we get two results,
$$A+B+C+D\ge 4\cdot\sqrt[4]{ABCD}$$
and
$$4-(A+B+C+D)\ge 4\cdot\sqrt[4]{\smash[b]{(1-A)(1-B)(1-C)(1-D)}}$$
Can I somehow use both these results to prove the inequality? Please help
 [/INST] By AM-GM we obtain:
$$\sqrt[4]{\prod\limits_{cyc}(1-a^4)}\geq\sqrt[4]{\prod\limits_{cyc}((a+b+c+d)^4-a^4)}=$$
$$=\sqrt[4]{\prod_{cyc}\left((b+c+d)\left((a+b+c+d)^3+(a+b+c+d)^2a+(a+b+c+d)a^2+a^3\right)\right)}\geq$$
$$\geq\sqrt[4]{\prod_{cyc}\left(3\sqrt[3]{bcd}\left(64\sqrt[4]{a^3b^3c^3d^3}+16a\sqrt{abcd}+4\sqrt[4]{abcd}a^2+a^3\right)\right)}\geq$$
$$\geq\sqrt[4]{\prod_{cyc}\left(3\sqrt[3]{bcd}\cdot85\sqrt[85]{(abcd)^{48}\cdot a^{16}\cdot(abcd)^8\cdot(abcd)\cdot a^8\cdot a^3}\right)}=$$
$$=255\sqrt[4]{\prod_{cyc}a^{\frac{84}{85}}b^{\frac{256}{255}}c^{\frac{256}{255}}d^{\frac{256}{255}}}=255abcd.$$
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