For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$ As in the title. Let $x,y$ be two real numbers such that $x+y=1$. Prove that $x^4+y^4\ge \frac{1}{8}$.
Any hints? Basically, the only method I am aware of is plugging $y=1-x$ into the inequality and investigating the extrema of the function, but I don't think it's the best method. I'm looking for a cleverer way to prove that inequality.

$$(x+y)^4=1 \\
x^4+4x^3y+6x^2y^2+4x^3+y^4 =1$$
Now, by AM-GM we have
$$x^3y \leq \frac{x^4+x^4+x^4+y^4}{4}\\
x^2y^2 \leq \frac{x^4+y^4}{2}\\
xy^3 \leq \frac{x^4+y^4+y^4+y^4}{4}\\
$$
Thus
$$1=x^4+4x^3y+6x^2y^2+4x^3+y^4  \leq x^3+(3x^4+y^4)+3(x^4+y^4)+(x^4+3y^4)+y^4$$
