Inegalitate conditionata de ab+bc+ca=0

[Claudiu Mindrila]

Fie \( a,\ b,\ c\in\mathbb{R}^{*} \) astfel incat \( ab+bc+ca=0 \).

a) Verificati ca \( \frac{bc\left(a-1\right)}{a}+\frac{ca\left(b-1\right)}{b}+\frac{ab\left(c-1\right)}{c}=2\left(a+b+c\right) \).
b) Aratati ca \( \frac{4}{3}\left(a^{2}+b^{2}+c^{2}\right)\ge a^{2}\left(b-1\right)\left(c-1\right)+b^{2}\left(c-1\right)\left(a-1\right)+c^{2}\left(a-1\right)\left(b-1\right) \).

Claudiu Mindrila, R. M. T. 1/2010

[marius00]

a) am sarit peste etapele evidente
\(
\[
- \frac{{bc}}{a} - \frac{{ac}}{b} - \frac{{ab}}{c} = - (\frac{{b^2 c^2 + a^2 c^2 + a^2 b^2 }}{{abc}}) = - [ - \frac{{(2a^2 bc + 2ab^2 c + 2abc^2 )}}{{abc}}] = 2(a + b + c)
\]
\)

[Andi Brojbeanu]

b)\( LHS=\frac{4}{3}(a^2+b^2+c^2)=\frac{4[(a+b+c)^2-2(ab+bc+ca)]}{3}=\frac{4(a+b+c)^2}{3}=\frac{[2(a+b+c)^2]}{3}=\frac{[\frac{bc(a-1)}{a}+\frac{ca(b-1)}{b}+\frac{ab(c-1)}{c}]^2}{3}\ge \)
\( \ge \frac{bc(a-1)}{a}\cdot \frac{ca(b-1)}{b}+\frac{ca(b-1)}{b}\cdot \frac{ab(c-1)}{c}+\frac{ab(c-1)}{c}\cdot \frac{bc(a-1)}{a}=a^2(b-1)(c-1)+b^2(c-1)(a-1)+c^2(a-1)(b-1)=RHS \).