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

$$4-(A+B+C+D)\ge 4\cdot\sqrt[4]{\smash[b]{(1-A)(1-B)(1-C)(1-D)}}$$

By AM-GM we obtain: $$\sqrt[4]{\prod\limits_{cyc}(1-a^4)}\geq\sqrt[4]{\prod\limits_{cyc}((a+b+c+d)^4-a^4)}=$$ $$=\sqrt[4]{\prod_{cyc}\left((b+c+d)\left((a+b+c+d)^3+(a+b+c+d)^2a+(a+b+c+d)a^2+a^3\right)\right)}\geq$$ $$\geq\sqrt[4]{\prod_{cyc}\left(3\sqrt[3]{bcd}\left(64\sqrt[4]{a^3b^3c^3d^3}+16a\sqrt{abcd}+4\sqrt[4]{abcd}a^2+a^3\right)\right)}\geq$$ $$\geq\sqrt[4]{\prod_{cyc}\left(3\sqrt[3]{bcd}\cdot85\sqrt[85]{(abcd)^{48}\cdot a^{16}\cdot(abcd)^8\cdot(abcd)\cdot a^8\cdot a^3}\right)}=$$ $$=255\sqrt[4]{\prod_{cyc}a^{\frac{84}{85}}b^{\frac{256}{255}}c^{\frac{256}{255}}d^{\frac{256}{255}}}=255abcd.$$