Inegalitate conditionata cu o inegalitate

[Marius Mainea]

Fie a,b,c>0, astfel incat \( a+b+c\leq\frac{3}{2} \). Demonstrati ca:

\( \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(b+a)^2}\geq\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \)

[Claudiu Mindrila]

Presupunem din cauza simetriei, ca \( a \leq b \leq c \). Tripletele \( (\frac{a}{b+c}, \frac{b}{a+c},\frac{c}{a+b}) \) si \( (\frac{1}{b+c}, \frac{1}{a+c}, \frac{1}{a+b}) \) sunt la fel ordonate, motiv pentru care putem aplica inegalitatea lui Cebisev.
Obtinem ca \( \Sigma \frac{a}{b+c} \cdot \Sigma\frac{1}{b+c} \leq 3 \cdot\Sigma \frac{a}{(b+c)^2}.\ (1) \)
Tinand cont ca \( a+b+c \leq \frac{3}{2} \), din inegalitatea Cauchy-Buniakowski-Schwarz obtinem ca \( \Sigma \frac{1}{b+c} \geq \frac{9}{2(a+b+c)} \geq \frac{9}{2} \cdot \frac{2}{3}=3.\ (2) \)
Din \( (1) \) si \( (2) \) rezulta ca \( 3 \cdot \Sigma \frac{a}{(b+c)^2} \geq 3 \cdot \Sigma \frac{a}{b+c} \), ceea ce trebuia aratat.