Doua inegalitati din Arhimede

[Claudiu Mindrila]

\( 1. \)Fie \( a,b,c \in (0, \infty) \) si daca \( (abc)^3=2 \), sa se arate ca: \( \frac{1}{a^3(1+b^3)}+\frac{1}{b^3(1+c^3)}+\frac{1}{c^3(1+a^3)} \geq 1 \)
Marcel Chirita, Arhimede 5-8/2005

\( 2. \) Daca \( a,b,c \in [0,1] \), atunci are loc inegalitatea: \( \frac{a}{4+b^2+c^2}+\frac{b}{4+c^2+a^2}+\frac{c}{4+a^2+b^2} \leq \frac{1}{2} \)
Virgil Nicula, Arhimede 9-12/2005

[BogdanCNFB]

2) Avem \( \sum\frac{a}{4+b^2+c^2}\leq\sum\frac{a}{4+2bc}=\frac{1}{2}\cdot\sum\frac{a}{2+bc}\leq\frac{1}{2}\cdot\sum\frac{a}{1+b+c}\leq\frac{1}{2}\cdot\sum\frac{a}{a+b+c}=\frac{1}{2} \).
Am folosit faptul ca \( a,b\in[0,1]\Leftrightarrow (1-a)(1-b)\ge 0\Leftrightarrow 1+ab\ge a+b \)

[Claudiu Mindrila]

Sau...deoarece \( 4+b^2+c^2 \geq 2b+2c+2 \geq 2(a+b+c) \) rezulta ca \( \sum \frac{a}{4+b^2+c^2} \leq \frac{\sum a}{2(a+b+c)}=\frac{1}{2} \). Rezulta cerinta...