Daca \( a,\ b,\ c\in\mathbb{R}_{+}^{*} \) si \( a+b+c=1 \), demonstrati inegalitatea
\( 3+\sum\frac{\left(a-b\right)^{2}+\left(a-c\right)^{2}}{1+a}\le4\left(a^{2}+b^{2}+c^{2}\right)\left(\sum\frac{1}{1+a}\right) \).
Titu Zvonaru, RecMat 1/2009
Inegalitate cu variabila pozitive
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Inegalitate cu variabila pozitive
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
Inegalitatea e echivalenta cu :
\( \sum\limits_{cyc} {\frac {{1 + a + 2a^2 + b^2 + c^2 - 2a\left( {b + c} \right)}}{{1 + a}}} \le 4\left( {a^2 + b^2 + c^2 } \right)\left( {\sum\limits_{cyc} {\frac {1}{{a + 1}}} } \right) \)
\( \Leftrightarrow \sum\limits_{cyc} {\frac {{4\left( {a^2 + b^2 + c^2 } \right) - \left( {1 + a + 2a^2 + b^2 + c^2 - 2a\left( {b + c} \right)} \right)}}{{a + 1}}} \ge 0 \)
\( \Leftrightarrow \sum\limits_{cyc} {\frac {{a - 1 + 3\left( {b^2 + c^2 } \right)}}{{1 + a}}} \ge 0 \)
Folosind \( b^2 + c^2 \ge \frac {{\left( {b + c} \right)^2 }}{2} \) si \( a + b + c = 1 \) e de ajuns sa demonstram ca :
\( \sum\limits_{cyc} {\frac {{\left( {b + c} \right)^2 }}{{1 + a}} \ge \sum\limits_{cyc} {\frac {{2a\left( {b + c} \right)}}{{1 + a}}} } \)
Adevarat deoarece \( \sum\limits_{cyc} {\frac {{\left( {b + c} \right)^2 }}{{1 + a}}} \ge \frac {{\left( {2\left( {a + b + c} \right)} \right)^2 }}{{3 + a + b + c}} = 1 \)
si
\( \sum\limits_{cyc} {\frac {{2a\left( {b + c} \right)}}{{1 + a}} \le \sum\limits_{cyc} {\frac {{\left( {2a + b + c} \right)^2 }}{{4\left( {1 + a} \right)}} = } } \sum\limits_{cyc} {\frac {{1 + a}}{4} = 1} \)
\( \sum\limits_{cyc} {\frac {{1 + a + 2a^2 + b^2 + c^2 - 2a\left( {b + c} \right)}}{{1 + a}}} \le 4\left( {a^2 + b^2 + c^2 } \right)\left( {\sum\limits_{cyc} {\frac {1}{{a + 1}}} } \right) \)
\( \Leftrightarrow \sum\limits_{cyc} {\frac {{4\left( {a^2 + b^2 + c^2 } \right) - \left( {1 + a + 2a^2 + b^2 + c^2 - 2a\left( {b + c} \right)} \right)}}{{a + 1}}} \ge 0 \)
\( \Leftrightarrow \sum\limits_{cyc} {\frac {{a - 1 + 3\left( {b^2 + c^2 } \right)}}{{1 + a}}} \ge 0 \)
Folosind \( b^2 + c^2 \ge \frac {{\left( {b + c} \right)^2 }}{2} \) si \( a + b + c = 1 \) e de ajuns sa demonstram ca :
\( \sum\limits_{cyc} {\frac {{\left( {b + c} \right)^2 }}{{1 + a}} \ge \sum\limits_{cyc} {\frac {{2a\left( {b + c} \right)}}{{1 + a}}} } \)
Adevarat deoarece \( \sum\limits_{cyc} {\frac {{\left( {b + c} \right)^2 }}{{1 + a}}} \ge \frac {{\left( {2\left( {a + b + c} \right)} \right)^2 }}{{3 + a + b + c}} = 1 \)
si
\( \sum\limits_{cyc} {\frac {{2a\left( {b + c} \right)}}{{1 + a}} \le \sum\limits_{cyc} {\frac {{\left( {2a + b + c} \right)^2 }}{{4\left( {1 + a} \right)}} = } } \sum\limits_{cyc} {\frac {{1 + a}}{4} = 1} \)
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