Inegalitate conditionata

[Radu Titiu]

Daca Fie a,b,c numere reale strict pozitive a.i. \( a^3+b^3+3c=5 \). Aratati ca:
\( \sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \)

[Marius Dragoi]

O idee, ceva...

[Radu Titiu]

Daca Fie a,b,c numere reale strict pozitive a.i. \( a^3+b^3+3c=5 \). Aratati ca:
\( \sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \)

Din inegalitatea Cauchy avem \( \sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\leq \sqrt{2(a+b+c)\left( \frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\right)} \leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \Leftrightarrow \)
\( a+b+c \leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \)(*)

Dar din conditia initiala avem :\( 9=(a^3+2)+(b^3+2)+3c\geq 3(a+b+c) \)
Deci \( a+b+c \leq 3 \)
Mai mult din inegalitatea Cauchy avem de asemenea :\( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}\geq 3 \)
Asadar relatia (*) este adevarata deoarece \( a+b+c \leq 3\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \)