<s>[INST] 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.
 [/INST] Also a Cauchy-Schwarz approach works:
\begin{align*}
1 = x + y \le \sqrt{(x^2+y^2)(1^2+1^2)} \implies x^2 + y^2 \ge \frac{1}{2} \\
\frac{1}{2} \le x^2 + y^2 \le \sqrt{(x^4 + y^4)(1^2 + 1^2)} \implies x^4 + y^4 \ge \frac{1}{8}
\end{align*}
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