mateforum.ro

Sa se arate ca:

\( \frac{a-d}{b+c+2d}+\frac{b-a}{c+d+2a}+\frac{c-b}{d+a+2b}+\frac{d-c}{a+b+2c}\ge 0 \)

oricare ar fi numerele reale pozitive a, b, c, d.

V. Berghea, G.M.

Inegalitatea din enunt este echivalenta cu \( \frac{a}{b+c+2d}+\frac{b}{c+d+2a}+\frac{c}{d+a+2b}+\frac{d}{a+b+2c}\ge\frac{d}{a+b+2c}+\frac{a}{c+d+2a}+\frac{b}{d+a+2b}+\frac{c}{a+b+2c} \).

Adunand
\( \frac{b+c+d}{b+c+2d}+\frac{c+d+a}{c+d+2a}+\frac{d+a+b}{d+a+2b}+\frac{a+b+c}{a+b+2c} \)
in ambii membri avem ca
\( \left(a+b+c+d\right)\left(\frac{1}{b+c+2d}+\frac{1}{c+d+2a}+\frac{1}{d+a+2b}+\frac{1}{a+b+2c}\right)\ge4 \)
sau
\( \frac{1}{b+c+2d}+\frac{1}{c+d+2a}+\frac{1}{d+a+2b}+\frac{1}{a+b+2c}\ge\frac{4}{a+b+c+d} \).

Aceasta din urma este inegalitatea Cauchy-Buniakowski-Schwarz:
\( \frac{1}{b+c+2d}+\frac{1}{c+d+2a}+\frac{1}{d+a+2b}+\frac{1}{a+b+2c}\ge\frac{4^{2}}{4\left(a+b+c+d\right)}=\frac{4}{a+b+c+d}. \)