Problema 1, al 3-lea Baraj pentru Juniori, 2009

[Claudiu Mindrila]

Fie \( a,\ b,\ c \) trei numere reale strict pozitive astfel incat \( a+b+c\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \). Sa se arate ca \( \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \).

[Marius Mainea]

Se foloseste lema:

Daca m,n,p,x,y,z sunt pozitive astfel incat \( mx+ny+pz\le m+n+p \), atunci \( \frac{m}{x}+\frac{n}{y}+\frac{p}{z}\ge m+n+p \).

[Claudiu Mindrila]

Problema revine la a demonstra inegalitatea: \( \sum a^{2}c\ge\sum a \).

Dar, conform ipotezei si inegalitatii Cauchy-Buniakowski-Schwarz avem:
\( \sum a\ge\sum\frac{1}{a}\Longrightarrow\left(\sum a\right)\left(\sum a^{2}c\right)\ge\left(\sum\frac{1}{a}\right)\left(\sum a^{2}c\right){}_{\ge}^{CBS}\left(\sum\sqrt{\frac{1}{a}\cdot b^{2}a}\right)^{2}=\left(\sum b\right)^{2} \), de unde este evident ca \( \sum a^{2}c\ge\sum a \).

[Claudiu Mindrila]

Solutia a 2-a: Conform ipotezei avem \( \sum a\ge\sum\frac{1}{a}\Longrightarrow abc\left(\sum a\right)\ge\sum ab \).

Cu inegalitatea C.B.S. avem: \( \sum\frac{a}{b}=\sum\frac{a^{2}}{ab}\ge\frac{\left(\sum a\right)^{2}}{\sum ab} \).

Problema revine la \( \sum\frac{a}{b}\ge\sum ab\Longleftrightarrow\frac{\left(\sum a\right)^{2}}{\sum ab}\ge\sum\frac{1}{ab}=\frac{\sum c}{abc}\Longleftrightarrow abc\left(\sum a\right)\ge\sum ab \), adevarat.