Question-207834

Question Number 207834 by efronzo1 last updated on 28/May/24

\underset{⏟}{}

Answered by Berbere last updated on 28/May/24

a^{505} = x; y = b^{505}

\Leftrightarrow {\begin{cases} x + y . \\ m a x {x^{4} + y^{4}} \end{cases}

x, y \in {[0, 1]}^{2} \Rightarrow x^{4} ⩽ x; y^{4} < y \Rightarrow x^{4} + y^{4} ⩽ x + y = 1

x = 1, y = 0 x^{4} + y^{4} = 1

m a x (x^{4} + y^{4}) = 1

m i n u s i n g \sqrt[4]{\frac{x^{4} + y^{4}}{2}} ⩾ \frac{x + y}{2} = \frac{1}{2}

\Rightarrow x^{4} + y^{4} ⩾ \frac{1}{8} (x = \frac{1}{2}; y = \frac{1}{2})

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