Arătaţi că, dacă \( a,b,c\in [0, 1] \), atunci
\( \frac{a}{10+b^3+c^4}+\frac{b}{10+c^3+a^4}+\frac{c}{10+a^3+b^4}\le\frac{1}{4} \)
Evaluare in Educatie la Matematica 13-06-2009
Inegalitate cu variabile subunitare 2
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Marius Mainea
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Claudiu Mindrila
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Re: Inegalitate cu variabile subunitare 2
Avem
\( \sum\frac{a}{10+b^{3}+c^{4}}\le\sum\frac{a}{10+b^{4}+c^{4}}\le\sum\frac{a}{9+a^{4}+b^{4}+c^{4}}=\frac{a+b+c}{9+a^{4}+b^{4}+c^{4}} \le \frac{1}{4} \).
Aceasta din urma este adevarata, deoarece \( 9+\sum a^{4}=\sum\left(a^{4}+3\right) \ge \sum4a=4\sum a \).
\( \sum\frac{a}{10+b^{3}+c^{4}}\le\sum\frac{a}{10+b^{4}+c^{4}}\le\sum\frac{a}{9+a^{4}+b^{4}+c^{4}}=\frac{a+b+c}{9+a^{4}+b^{4}+c^{4}} \le \frac{1}{4} \).
Aceasta din urma este adevarata, deoarece \( 9+\sum a^{4}=\sum\left(a^{4}+3\right) \ge \sum4a=4\sum a \).
Last edited by Claudiu Mindrila on Sat Jun 13, 2009 3:51 pm, edited 1 time in total.
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Andi Brojbeanu
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\( \frac{a}{10+b^3+c^4}\leq \frac{a}{4(a+b+c)}\Leftrightarrow 4a+4b+4c\leq 10+b^3+c^4 \).
\( 10+b^3+c^4=4\cdot 1+ (b^3+1+1)+(c^4+1+1+1)+1\geq 4a+3\sqrt[3]{b^3\cdot 1\cdot 1}+4\sqrt[4]{c^4\cdot 1\cdot 1\cdot 1}+ b=4a+4b+4c \).
Deci, \( \sum\frac{a}{10+b^3+c^4}\leq \frac{a+b+c}{4(a+b+c)}=\frac{1}{4} \).
\( 10+b^3+c^4=4\cdot 1+ (b^3+1+1)+(c^4+1+1+1)+1\geq 4a+3\sqrt[3]{b^3\cdot 1\cdot 1}+4\sqrt[4]{c^4\cdot 1\cdot 1\cdot 1}+ b=4a+4b+4c \).
Deci, \( \sum\frac{a}{10+b^3+c^4}\leq \frac{a+b+c}{4(a+b+c)}=\frac{1}{4} \).