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If $$x$$ and $$y$$ are positive real numbers such that $$x+y=1$$, then the maximum value of $$xy^4+x^4y$$ is
Let $$t=xy$$. Then $$xy^4+x^4y=xy(x^3+y^3)=t(1-3t)$$ because $$x+y=1$$. Since $$0<t\le\frac14$$, the quadratic $$t-3t^2$$ is maximized at $$t=\frac16$$, giving the maximum value $$\frac{1}{12}$$.
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