Inegalitate conditionata

[Marius Mainea]

Daca a,b,c sunt numere pozitive cu abc=1, atunci

\( \frac{a^2+b^2}{a^2+b^2+1}+\frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1}\ge \frac{a+b}{a^2+b^2+1}+\frac{b+c}{b^2+c^2+1}+\frac{c+a}{c^2+a^2+1} \)

Jingjun Han, Mathematical Reflections 6/2008

[Claudiu Mindrila]

Deoarece \( \frac{a^{2}+b^{2}}{a^{2}+b^{2}+1}=1-\frac{1}{a^{2}+b^{2}+1} \) problema se reduce la a demonstra inegalitatea \( \sum\frac{a+b+1}{a^{2}+b^{2}+1}\leq3. \) Dar cum \( a^{2}+b^{2}+1\geq\frac{\left(a+b+1\right)^{2}}{3} \) avem \( \sum\frac{a+b+1}{a^{2}+b^{2}+1}\leq\sum\frac{3\left(a+b+1\right)}{\left(a+b+1\right)^{2}}=3\cdot\sum\frac{1}{a+b+1}. \) Mai ramane de demonstrat ca \( \sum\frac{1}{a+b+1}\leq1. \). Fie \( a=x^{3},b=y^{3},c=z^{3} \) cu \( x,y,z\geq0 \) si \( xyz=1. \). Rezulta ca
\( \sum\frac{1}{a+b+1}=\sum\frac{1}{x^{3}+y^{3}+xyz}\leq\sum\frac{1}{xy\left(x+y\right)+xyz}=\sum\frac{1}{xy\left(x+y+z\right)}=\sum\frac{z}{x+y+z}=1 \qed \).
Observatie. Pentru ultima inegalitate am folosit faptul ca pentru orice \( m,n \in \mathbb{R} \) avem \( m^3+n^3 \geq mn(m+n) \left(\Longleftrightarrow (m-n)^2(m+n) \geq 0 \right ) \)

[baleanuAR]

http://reflections.awesomemath.org/2008 ... utions.pdf