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Sa se arate ca \( \left\{\ \begin{array}{c}
x > y > 0\\\
a > b > 0\end{array}\ \right\|\ \Longrightarrow\ \left(x^b - y^b\right)^a < \left(x^a - y^a\right)^b \) .

Notam \( \frac{x}{y}=\alpha>1 \).Inegaliatea se rescrie
\( \left({\alpha}^b - 1\right)^a < \left({\alpha}^a - 1\right)^b . \)care devine succesiv
\( \left({\alpha}^b - 1\right)^{\frac{a}{b}} < \left({\alpha}^a - 1\right) . \)
\( \left({\alpha}^b - 1\right)^{\frac{a}{b}}+1 < {\alpha}^a \)
Notam \( \frac{a}{b}=\beta>1 \) si consideram functia
\( f: (0,\infty) \to R,f(x)=(x+1)^\beta-x^\beta-1 \).Prin derivare avem ca
\( f^\prime(x)=\beta (x+1)^{\beta-1}-\beta x^{\beta-1} \).
Cum \( \beta-1>0 \) si \( x+1>x \),rezulta ca \( f^\prime(x) \) este pozitiva,deci f crescatoare.
Deci \( f(x)>f(0)=0 \),iar pentru \( x=\alpha^b-1 \),obtinem inegalitatea dorita.

P.S.:Cred ca o alta solutie ar putea aparea daca notam \( f(x)=x^b,g(x)=x^a \) si \( h: (y,\infty) \to R, h(x)=f(g(x)-g(y))-g(f(x)-f(y)) \) si studiem daca h are semn constant pe \( (y,\infty) \)(h fiind continua).