alta inegalitate ONM SHL 2004

[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

[Marius Mainea]

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 \)

[Cezar Lupu]

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...