Fie a,b,c numere reale pozitive astfel incat \( a+b+c\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}. \) Demonstrati inegalitatea:
\( \sum {\frac{a^3c}{b(c+a)}\geq \frac{3}{2}. \)
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Fie a, b, c ...
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Marius Mainea
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Din inegalitatea lui Hölder stim ca
\( \sum\frac{a^3c}{b(c+a)}\sum(c+a)\sum\frac{b}{c}\geq\(\sum a\)^3. \)
Prin urmare \( \sum\frac{a^3c}{b(c+a)}\geq\frac{\(\sum a\)^3}{2\sum a\sum\frac{b}{c}}. \) Dar, din ipoteza \( \sum\frac{b}{c}\leq\sum a \) deci \( \sum\frac{a^3c}{b(c+a)}\geq\frac{\sum a}{2}\geq\frac{\sum\frac{b}{c}}{2}\geq\frac{3}{2}, \) ultima inegalitate fiind obtinuta din medii.
\( \sum\frac{a^3c}{b(c+a)}\sum(c+a)\sum\frac{b}{c}\geq\(\sum a\)^3. \)
Prin urmare \( \sum\frac{a^3c}{b(c+a)}\geq\frac{\(\sum a\)^3}{2\sum a\sum\frac{b}{c}}. \) Dar, din ipoteza \( \sum\frac{b}{c}\leq\sum a \) deci \( \sum\frac{a^3c}{b(c+a)}\geq\frac{\sum a}{2}\geq\frac{\sum\frac{b}{c}}{2}\geq\frac{3}{2}, \) ultima inegalitate fiind obtinuta din medii.
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Cezar Lupu
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Alta solutie.
Avem \( \frac{a^3c}{b(c+a)}=\frac{\left(\frac{a}{b}\right)^{2}}{\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}\right)} \) plus analoagele.
Astfel, vom avea ca \( \sum_{cyc}\frac{a^3c}{b(c+a)}=\sum_{cyc}\frac{\left(\frac{a}{b}\right)^{2}}{\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}\right)} \).
Aplicand inegalitatea Cauchy-Schwarz, vom obtine
\( \left(\sum_{cyc}\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}\right)\right)\left(\sum_{cyc}\frac{\left(\frac{a}{b}\right)^{2}}{\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}\right)}\right)\geq\left(\sum_{cyc}\frac{a}{b}\right)^{2} \).
Astfel, vom avea ca
\( \sum_{cyc}\frac{\left(\frac{a}{b}\right)^{2}}{\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}\right)}\geq\frac{\left(\sum_{cyc}\frac{a}{b}\right)\left(\sum_{cyc}\frac{a}{b}\right)}{2\sum_{cyc}\frac{1}{ab}}. \)
Tot ce ne ramane acum, este sa demonstram ca \( \sum_{cyc}\frac{a}{b}\geq\sum_{cyc}\frac{1}{ab} \)
Presupunem prin reducere la absurd ca \( \sum_{cyc}\frac{a}{b}<\sum_{cyc}\frac{1}{ab} \) care este echivalenta cu \( \sum_{cyc}ab^2<\sum_{cyc}a \). Adunand aceasta ultima inegalitate cu inegalitatea din ipoteza scrisa sub forma \( 2\sum_{cyc}a\geq 2\sum_{cyc}\frac{a}{b} \), va rezulta ca
\( 3\sum_{cyc}a>a\left(b^2+\frac{2}{b}\right)+b\left(c^2+\frac{2}{c}\right)+c\left(a^2+\frac{2}{a}\right) \)
si tinand cont de inegalitatea mediilor sub forma \( x^2+\frac{2}{x}=x^2+\frac{1}{x}+\frac{1}{x}\geq 3\sqrt[3]{x^2\cdot\frac{1}{x}\cdot\frac{1}{x}}=3 \), rezulta contradictia \( 3(a+b+c)>3(a+b+c) \). \( \qed \)
Avem \( \frac{a^3c}{b(c+a)}=\frac{\left(\frac{a}{b}\right)^{2}}{\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}\right)} \) plus analoagele.
Astfel, vom avea ca \( \sum_{cyc}\frac{a^3c}{b(c+a)}=\sum_{cyc}\frac{\left(\frac{a}{b}\right)^{2}}{\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}\right)} \).
Aplicand inegalitatea Cauchy-Schwarz, vom obtine
\( \left(\sum_{cyc}\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}\right)\right)\left(\sum_{cyc}\frac{\left(\frac{a}{b}\right)^{2}}{\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}\right)}\right)\geq\left(\sum_{cyc}\frac{a}{b}\right)^{2} \).
Astfel, vom avea ca
\( \sum_{cyc}\frac{\left(\frac{a}{b}\right)^{2}}{\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}\right)}\geq\frac{\left(\sum_{cyc}\frac{a}{b}\right)\left(\sum_{cyc}\frac{a}{b}\right)}{2\sum_{cyc}\frac{1}{ab}}. \)
Tot ce ne ramane acum, este sa demonstram ca \( \sum_{cyc}\frac{a}{b}\geq\sum_{cyc}\frac{1}{ab} \)
Presupunem prin reducere la absurd ca \( \sum_{cyc}\frac{a}{b}<\sum_{cyc}\frac{1}{ab} \) care este echivalenta cu \( \sum_{cyc}ab^2<\sum_{cyc}a \). Adunand aceasta ultima inegalitate cu inegalitatea din ipoteza scrisa sub forma \( 2\sum_{cyc}a\geq 2\sum_{cyc}\frac{a}{b} \), va rezulta ca
\( 3\sum_{cyc}a>a\left(b^2+\frac{2}{b}\right)+b\left(c^2+\frac{2}{c}\right)+c\left(a^2+\frac{2}{a}\right) \)
si tinand cont de inegalitatea mediilor sub forma \( x^2+\frac{2}{x}=x^2+\frac{1}{x}+\frac{1}{x}\geq 3\sqrt[3]{x^2\cdot\frac{1}{x}\cdot\frac{1}{x}}=3 \), rezulta contradictia \( 3(a+b+c)>3(a+b+c) \). \( \qed \)