O inegalitate tot de clasa a cincea
Moderators: Bogdan Posa, Laurian Filip
O inegalitate tot de clasa a cincea
Demonstrati ca \( \frac{a+2^{180}}{a+3^{120}}<\frac{b+5^{900}}{b+7^{720}} \), oricare ar fi a si b numere naturale.
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Andi Brojbeanu
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\( 2^{180}=(2^3)^{60}=9^{60}; 3^{120}=(3^2)^{60}=8^{60};
5^{900}=(5^5)^{180}=3125^{180}; 7^{720}=(7^4)^{180}=2401^{180} \).
Inmultind mezii cu extremii, obtinem \( a\cdot b+a\cdot 2401^{180}+8^{60}\cdot b+8^{60}\cdot 2401^{180}<a\cdot b+a\cdot 3125^{180}+9^{60}\cdot b+9^{60}\cdot 3125^{180} \)
sau \( 0<a(3125^{180}-2401^{180})+b(9^{60}-8^{60})+9^{60}\cdot 3125^{180}-8^{60}\cdot 2401^{180}(1) \).
Dar, \( 3125^{180}>2401^{180}\Rightarrow a(3125^{180}-2401^{180})\geq 0(2) \) si \( 9^{60}>8^{60}\Rightarrow b(9^{60}-8^{60})\geq 0(3) \).
De asemenea, inmultind cele doua relatii, obtinem \( 9^{60}\cdot 3125^{180}>8^{60}\cdot 2401^{180}\Rightarrow 9^{60}\cdot 3125^{180}-8^{60}\cdot 2401^{180}>0(4) \).
Adunand relatiile \( (2), (3) si (4) \) obtinem relatia \( (1) \), care este adevarata, ceea ce trebuia demonstrat.
5^{900}=(5^5)^{180}=3125^{180}; 7^{720}=(7^4)^{180}=2401^{180} \).
Inmultind mezii cu extremii, obtinem \( a\cdot b+a\cdot 2401^{180}+8^{60}\cdot b+8^{60}\cdot 2401^{180}<a\cdot b+a\cdot 3125^{180}+9^{60}\cdot b+9^{60}\cdot 3125^{180} \)
sau \( 0<a(3125^{180}-2401^{180})+b(9^{60}-8^{60})+9^{60}\cdot 3125^{180}-8^{60}\cdot 2401^{180}(1) \).
Dar, \( 3125^{180}>2401^{180}\Rightarrow a(3125^{180}-2401^{180})\geq 0(2) \) si \( 9^{60}>8^{60}\Rightarrow b(9^{60}-8^{60})\geq 0(3) \).
De asemenea, inmultind cele doua relatii, obtinem \( 9^{60}\cdot 3125^{180}>8^{60}\cdot 2401^{180}\Rightarrow 9^{60}\cdot 3125^{180}-8^{60}\cdot 2401^{180}>0(4) \).
Adunand relatiile \( (2), (3) si (4) \) obtinem relatia \( (1) \), care este adevarata, ceea ce trebuia demonstrat.
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Laurian Filip
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