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SHL 2004
Posted: Mon Mar 22, 2010 8:51 pm
by Claudiu Mindrila
Daca \( a,\ b,\ c\ge0 \) atunci \( \sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+\left(a+b+c\right)^{2}\ge4\sqrt{3abc\left(a+b+c\right)} \).
Valentin Vornicu, lista scurta, 2004
Re: SHL 2004
Posted: Tue Mar 23, 2010 12:23 am
by Marius Mainea
Claudiu Mindrila wrote:Daca \( a,\ b,\ c\ge0 \) atunci \( \sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+\left(a+b+c\right)^{2}\ge4\sqrt{3abc\left(a+b+c\right)} \).
Valentin Vornicu, lista scurta, 2004
\( AM\ge GM \)
\( LHS=sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+\frac{(a+b+c)^2}{3}+\frac{(a+b+c)^2}{3}+\frac{(a+b+c)^2}{3}\ge 4\sqrt[4]{sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\(\frac{(a+b+c)^2}{3}\)^3}\ge RHS \)
Re: SHL 2004
Posted: Sat Mar 27, 2010 4:43 am
by Cezar Lupu
Marius Mainea wrote:Claudiu Mindrila wrote:Daca \( a,\ b,\ c\ge0 \) atunci \( \sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+\left(a+b+c\right)^{2}\ge4\sqrt{3abc\left(a+b+c\right)} \).
Valentin Vornicu, lista scurta, 2004
\( AM\ge GM \)
\( LHS=sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+\frac{(a+b+c)^2}{3}+\frac{(a+b+c)^2}{3}+\frac{(a+b+c)^2}{3}\ge 4\sqrt[4]{sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\(\frac{(a+b+c)^2}{3}\)^3}\ge RHS \)
Da, sau se poate folosi urmatoarea problema data la un baraj OIM in 2001, anume:
Daca \( a, b, c \) sunt numere reale strict pozitive, atunci are loc inegalitatea:
\( \sum_{cyc}(b+c-a)(c+a-b)\leq\sqrt{abc}{(\sqrt{a}+\sqrt{b}+\sqrt{c}) \).
Las pe cei interesati sa continue solutia mai departe...
