a=\sqrt2^{\sqrt3} , b=\sqrt3^{\sqrt2}
Niech
b>a
\sqrt3^{\sqrt2}>\sqrt2^{\sqrt3}
(3^{\frac{1}{2}})^{\sqrt2}>(2^{\frac{1}{2}})^{\sqrt{3}}
3^{\frac{\sqrt2}{2}}>2^{\frac{\sqrt3}{2}} \ |^2 obustronnie do kwadratu
3^{\sqrt2}>2^{\sqrt3} \ |^{\sqrt2}
3^2>2^{\sqrt6}
9>2^{\sqrt6}
L>P
bo 2^{\sqrt6}<2^{\sqrt9} \Rightarrow 2^{\sqrt6}<2^3 \Rightarrow 2^{\sqrt6}<8
co kończy dowód
Odpowiedź:
b>a