Question Number 207834 by efronzo1 last updated on 28/May/24
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Answered by Berbere last updated on 28/May/24
![a^(505) =x;y=b^(505) ⇔ { ((x+y.)),((max{x^4 +y^4 })) :} x,y∈[0,1]^2 ⇒x^4 ≤x;y^4 <y⇒x^4 +y^4 ≤x+y=1 x=1,y=0 x^4 +y^4 =1 max(x^4 +y^4 )=1 min using (((x^4 +y^4 )/2))^(1/4) ≥((x+y)/2)=(1/2) ⇒x^4 +y^4 ≥(1/8)(x=(1/2);y=(1/2))](Q207836.png)
$${a}^{\mathrm{505}} ={x};{y}={b}^{\mathrm{505}} \\ $$$$\Leftrightarrow\begin{cases}{{x}+{y}.}\\{{max}\left\{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right\}}\end{cases} \\ $$$${x},{y}\in\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} \Rightarrow{x}^{\mathrm{4}} \leqslant{x};{y}^{\mathrm{4}} <{y}\Rightarrow{x}^{\mathrm{4}} +{y}^{\mathrm{4}} \leqslant{x}+{y}=\mathrm{1} \\ $$$${x}=\mathrm{1},{y}=\mathrm{0}\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\mathrm{1} \\ $$$${max}\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)=\mathrm{1} \\ $$$${min}\:{using}\:\sqrt[{\mathrm{4}}]{\frac{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }{\mathrm{2}}}\geqslant\frac{{x}+{y}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +{y}^{\mathrm{4}} \geqslant\frac{\mathrm{1}}{\mathrm{8}}\left({x}=\frac{\mathrm{1}}{\mathrm{2}};{y}=\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$