Fie \( a,b,c>0 \) astfel incat \( abc=1 \) . Sa se demonstreze ca :
a)\( \frac{a-1}{b}+\frac{b-1}{c}+\frac{c-1}{a}\ge 0 \)
b)\( \frac{a-1}{b+c}+\frac{b-1}{c+a}+\frac{c-1}{a+b}\ge 0 \)
Doua inegalitati cu aceeiasi conditie
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Doua inegalitati cu aceeiasi conditie
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Mateescu Constantin
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a) Inegalitatea se scrie echivalent \( ab^2+bc^2+ca^2\ge a+b+c \)
Din \( AM-GM \) avem
\( 3(ab^2+bc^2+ca^2)=\sum(2ab^2+bc^2)\ge \sum3\sqrt[3]{a^2b^5c^2}=3(b+c+a) \)
b) Inegalitatea se scrie \( 3abc+a^3+b^3+c^3-a^2-b^2-c^2-3(ab+bc+ca)+a^2(b+c)+b^2(c+a)+c^2(a+b)=\sum a^3-\sum a^2+(a+b+c-3)(ab+bc+ca)\ge 0 \)
E clar ca \( a+b+c\ge 3 \), deci ramane de aratat ca \( \sum a^3-\sum a^2\ge 0 \)
Aplicand \( AM-GM \) obtinem
\( 9\sum a^3=\sum(7a^3+b^3+c^3)\ge \sum 9\sqrt[9]{a^21b^3c^3}=9\sum a^2 \).
Din \( AM-GM \) avem
\( 3(ab^2+bc^2+ca^2)=\sum(2ab^2+bc^2)\ge \sum3\sqrt[3]{a^2b^5c^2}=3(b+c+a) \)
b) Inegalitatea se scrie \( 3abc+a^3+b^3+c^3-a^2-b^2-c^2-3(ab+bc+ca)+a^2(b+c)+b^2(c+a)+c^2(a+b)=\sum a^3-\sum a^2+(a+b+c-3)(ab+bc+ca)\ge 0 \)
E clar ca \( a+b+c\ge 3 \), deci ramane de aratat ca \( \sum a^3-\sum a^2\ge 0 \)
Aplicand \( AM-GM \) obtinem
\( 9\sum a^3=\sum(7a^3+b^3+c^3)\ge \sum 9\sqrt[9]{a^21b^3c^3}=9\sum a^2 \).