Inegalitate draguta 2

[Claudiu Mindrila]

Daca \( x_1,x_2,x_3,x_4 \) sunt numere reale strict pozitive atunci:
\( \frac{x_1}{x_2+x_3}+\frac{x_2}{x_3+x_4}+\frac{x_3}{x_4+x_1}+\frac{x_4}{x_1+x_2} \geq
\)\( \frac{x_1}{x_1+x_2}+\frac{x_2}{x_2+x_3}+\frac{x_3}{x_3+x_4}+\frac{x_4}{x_4+x_1}. \)
Sorin Radulescu, Marius Radulescu, Concursul revistei "Arhimede", 2003

[Mateescu Constantin]

Daca \( a,\ b,\ c,\ d \) sunt numere reale strict pozitive atunci:
\( \frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a} \)

Sorin Radulescu, Marius Radulescu, Concursul revistei "Arhimede", 2003

Inegalitatea se scrie echivalent: \( \sum\frac{a}{b+c}\ge 4+\sum\left\(\frac{a}{a+b}-1 \right\)=4+\sum\frac{-b}{a+b} \)

\( \Longleftrightarrow\ \frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{a+c}{a+d}+\frac{b+d}{a+b}\geq 4 \)

\( \Longleftrightarrow\ \frac{(a+c)(a+b+c+d)}{(b+c)(a+d)}+\frac{(b+d)(a+b+c+d)}{(a+b)(c+d)}\geq4 \)

Dar din \( AM-GM \) avem \( 4(b+c)(a+d)\le (a+b+c+d)^2 \) si \( 4(a+b)(c+d)\le (a+b+c+d)^2 \)

\( \Longrightarrow\ \frac{(a+c)(a+b+c+d)}{(b+c)(a+d)}+\frac{(b+d)(a+b+c+d)}{(a+b)(c+d)}\ge\frac{4(a+c)(a+b+c+d)}{(a+b+c+d)^{2}}+\frac{4(b+d)(a+b+c+d)}{(a+b+c+d)^{2}}=4 \)