(M2) Subiectul II-faza finala-Concursul de Evaluare

[Laurian Filip]

Sa se arate ca
\( x^{\log_52}+x^{\log_59}+x^{\log_524}\geq x^{\log_53}+x^{\log_58}+x^{\log_518} \), \( \forall x \in (0,\infty). \)

[Tudor Micu]

Fie \( x^{\log_5 2}=a \) \( x^{\log_5 3}=b \).
Inegalitatea revine la:
\( a+b^2+a^3b\geq b+a^3+ab^2 \)
\( a^3b-b-a^3\geq ab^2-a-b^2 \)
\( (a^3-1)(b-1)\geq (a-1)(b^2-1) \)
\( (b-1)(a-1)(a^2+a-b)\geq 0 \)
Daca \( x\geq 1 \) avem \( a,b\geq 1 \), iar \( a^2+a-b=x^{\log_5 4}+x^{\log_5 2}-x^{\log_5 3}\geq 0 \)
Daca \( x<1 \) avem \( a,b<1 \).
Fie \( x=\frac{1}{v} \). Rezulta \( a^2+a-b=\frac{1}{v^{\log_5 4}}+\frac{1}{v^{\log_5 2}}-\frac{1}{v^{\log_5 3}} \)
Ramane de aratat ca \( \frac{1}{v^{\log_5 4}}+\frac{1}{v^{\log_5 2}}-\frac{1}{v^{\log_5 3}}\geq 0 \), echivalent cu \( 1+v^{\log_5 2}-v^{\log_5 \frac{4}{3}\geq 0 \)

O alta varianta asemanatoare:
\( \log_5 2=\frac{\log_x 2}{\log_x 5} \)
\( (x^{\log_x 2})^{\log_5 x}=2^{\log_5 x} \)
Inegalitatea revine deci la:
\( 2^y+9^y+24^y\geq 3^y+8^y+18^y \), unde \( y=\log_5 x \)
\( 2^y+9^y-2^y\cdot 9^y\geq 3^y+8^y-3^y\cdot 8^y \)
\( (2^y-1)(9^y-1)\leq (3^y-1)(8^y-1 \))
\( (2^y-1)(3^y-1)(2^{2y}+2^y-3^y)\geq 0 \)
De aici este ca la solutia precedenta.

[Laurian Filip]

O alta varianta asemanatoare:
\( \log_5 2=\frac{\log_x 2}{\log_x 5} \)
\( (x^{\log_x 2})^{\log_5 x}=2^{\log_5 x} \)
Inegalitatea revine deci la:
\( 2^y+9^y+24^y\geq 3^y+8^y+18^y \), unde \( y=\log_5 x \)
\( 2^y+9^y-2^y\cdot 9^y\geq 3^y+8^y-3^y\cdot 8^y \)
\( (2^y-1)(9^y-1)\leq (3^y-1)(8^y-1 \))
\( (2^y-1)(3^y-1)(2^{2y}+2^y-3^y)\geq 0 \)
De aici este ca la solutia precedenta.

Asa am facut-o si eu . E chiar banala dar m-am cam gandit ceva timp la ea.