Fie \( a,b,c \) numere reale strict pozitive care satisfac relatia \( a+b+c=ab+bc+ca \). Demonstrati ca are loc urmatoarea ineagalitate:
\( \sum_{cyc}\frac{a}{b+c}\leq \frac{9}{8}(a+b+c)-\frac{15}{8}. \)
Inegalitate "own"
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Inegalitate "own"
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Omogenizam inegalitatea :
\( (\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})(ab+bc+ca)\le\frac{9}{8}(a+b+c)^2-\frac{15}{8}(ab+bc+ca) \) si desfacand parantezele obtinem
\( \sum{\frac{abc}{b+c}}+\sum {\frac{a^2(b+c)}{b+c}}\le \frac{9}{8}\sum{a^2}-\frac{3}{8}\sum{ab} \) sau
\( \sum{\frac{abc}{b+c}}\le \frac{\sum{a^2}-3\sum{ab}}{8} \)
Dar \( \sum{\frac{abc}{b+c}}\le\sum{\frac{a(b+c)}{4}}=\frac{\sum{ab}}{2}\le\frac{\sum{a^2}-3\sum{ab}}{8} \)
\( \Longleftrightarrow\sum{ab}\le \sum{a^2} \) care este adevarat.
\( (\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})(ab+bc+ca)\le\frac{9}{8}(a+b+c)^2-\frac{15}{8}(ab+bc+ca) \) si desfacand parantezele obtinem
\( \sum{\frac{abc}{b+c}}+\sum {\frac{a^2(b+c)}{b+c}}\le \frac{9}{8}\sum{a^2}-\frac{3}{8}\sum{ab} \) sau
\( \sum{\frac{abc}{b+c}}\le \frac{\sum{a^2}-3\sum{ab}}{8} \)
Dar \( \sum{\frac{abc}{b+c}}\le\sum{\frac{a(b+c)}{4}}=\frac{\sum{ab}}{2}\le\frac{\sum{a^2}-3\sum{ab}}{8} \)
\( \Longleftrightarrow\sum{ab}\le \sum{a^2} \) care este adevarat.