<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] Apply Lagrangian multiplier method ! \begin{align}
\min \quad x^4 + y^4 \\ \text{s.t.} \quad x + y = 1 \end{align} gives $\mathcal{L}(x,y) = x^4 + y^4 - \lambda(x+y-1)$. FOC gives \begin{align} \mathcal{L}_x = 4x^3 - \lambda & = 0 \\ \mathcal{L}_y = 4y^3 - \lambda & = 0 \\ x+y & = 1. \end{align} Solving gives an optimum of $(x^*,y^*) = (\frac{1}{2},\frac{1}{2})$ with value $\frac{1}{8}$.
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