Question

Solve the inequation ${(x+3)}^4+{(x+5)}^4\ge 4$

Answer 1

${(x+3)}^4+{(x+5)}^4\ge 4$
Let $y=\frac{x+3+x+5}{2}=x+4$
So $x=y-4$
${(y-1)}^4+{(y+1)}^4\ge 4$

${(y^2+2y+1)}^2+{(y^2-2y+1)}^2\ge 4$

${(y}^4+4y^2+1+4y^3+4y+2y^2)+{(y}^4+4y^2+1-4y^3-4y+2y^2)\ge 4$

$y^4+4y^2+1+2y^2\ge 2$

$y^4+6y^2-1\ge 0$

$y^2=\frac{-6\pm \sqrt{10}}{2}$

$(y^2-(\frac{-6-\sqrt{10}}{2}\left)\right)(y^2-(\frac{-6+\sqrt{10}}{2}\left)\right)\ge 0$

So $x\in (-\infty ,-4-\sqrt{\sqrt{10}-3})\cup (-4+\sqrt{\sqrt{10}-3},\infty )$