Let $a_4\geq a_3\geq a_2\geq a_1$ and $a_4=a$, $a_3=b$, $a_2=c$, $a_1=d$ and $c=xd$.
Hence, $x\geq1$.
Let $a\geq b+c$, $b\geq c+d$ and $f(a)=2(a^4+b^4+c^4+d^4)-33abcd$.
Hence, $f'(a)=8a^3-33bcd\geq8(b+c)^3-33bcd>0$, which says that $$f(a)\geq f(b+c)=2(b+c)^4+2b^4+2c^4+2d^4-33(b+c)bcd$$ Let $g(b)=2(b+c)^4+2b^4+2c^4+2d^4-33(b+c)bcd$.
Hence, $g'(b)=8(b+c)^3+8b^3-66bcd-33c^2d$ and $g''(b)=24(b+c)^2+24b^3-66cd>0$,
which says that $g'(b)\geq8(2c+d)^3+8(c+d)^3-66(c+d)cd-33c^2d>0$,
which says that $$2(a^4+b^4+c^4+d^4)-33abcd\geq g(b)\geq g(c+d)=$$
$$=2(2c+d)^4+2(c+d)^4+2c^4+2d^4-33(2c+d)(c+d)cd=$$ $$=d^4\left(2(2x+1)^4+2(x+1)^4+2x^4+2-33(2x+1)(x+1)x\right)=$$ $$=3d^4(x-1)(3x+2)(4x^2+2x-1)\geq0$$ which is contradiction!
This says that $a
$b
Done!