mateforum.ro

\( 1. \)
Fie \( x,y,z\in \mathbb{R} \) cu \( x+y+z=1 \). Sa se arate ca:
\( x^2+y^2+z^2\geq 4 \cdot (xy+yz+zx)-1 \). In ce caz are loc egalitatea?
V. Branzanescu, O.N.M. 1990

\( 2. \)
Fie \( a,b,c \in [0,1] \) astfel incat \( ab+bc+ca=1. \) Demonstrati ca \( a^{2006}+b^{2006}+c^{2006} \leq 2 \).
Concursul "Cezar Ivanescu", 2006

\( 3. \)
Sa se arate ca \( \sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}<1 \), oricare ar fi numerele reale \( a,b,c \in (0,1) \).
Dinu Serbanescu, Test de Selectie pentru OBMJ, 2002

\( 4. \)
Fie \( a,b,c \) numere pozitive astfel incat \( a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}. \) Aratati ca \( a+b+c\geq \frac{3}{abc} \)
Cezar Lupu, Test de Selectie pentru OBMJ, 2005

\( 5. \)
Fie \( x,y,z \) numere reale pozitive cu proprietatea ca: \( \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=2. \) Sa se demonstreze ca \( 8xyz \leq 1 \).
Mircea Lascu, Test de Selectie pentru OBMJ, 2006

pt 5 este un articol interesant pe gazeta matematica :

http://www.gazetamatematica.net/?q=node/742

1. \( x^2+y^2+z^2+(x+y+z)^2\ge 4(xy+yz+zx) \)

3. \( a,\ b,\ c>0 \) \( \Longrightarrow \sqrt{abc}<\sqrt[3]{abc} \) si \( \sqrt{(1-a)(1-b)(1-c)}<\sqrt[3]{(1-a)(1-b)(1-c)} \)

Din \( AM-GM \)
\( \Longrightarrow \sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}<\sqrt[3]{abc}+\sqrt[3]{(1-a)(1-b)(1-c)}\le \frac{a+b+c}{3}+\frac{3-(a+b+c)}{3}=1. \)

4.

\( abc(a + b + c) \geq bc + ca + ab \geq \sqrt {3abc(a + b + c)} \)

Deci

\( abc(a + b + c) \geq 3 \)

3.

\( \sqrt {abc} + \sqrt {\left( 1 - a\right) \left( 1 - b\right) \left( 1 - c\right) } = \sqrt {a}\sqrt {bc} + \sqrt {1 - a}\sqrt {\left( 1 - b\right) \left( 1 - c\right) }\leq \sqrt {a + \left( 1 - a\right) }\sqrt {bc + \left( 1 - b\right) \left( 1 - c\right) } = \sqrt {bc + \left( 1 - b\right) \left( 1 - c\right) } = \sqrt {1 - \left( b + c\right) + 2bc} < 1 \)

4. Alta solutie:

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

1. Alta solutie:

Notam \( x-\frac{1}{3}=a \) si analoagele, de unde \( a+b+c=0 \).

Relatia de demonstrat devine:
\( \left\(a+\frac{1}{3}\right\)^2+\left\(b+\frac{1}{3}\right\)^2+\left\(c+\frac{1}{3}\right\)^2\ge 4\left\[\left\(a+\frac{1}{3}\right\)\left\(b+\frac{1}{3}\right\)+\left\(b+\frac{1}{3}\right \) \left\(c+\frac{1}{3}\right \)+\left\(c+\frac{1}{3}\right \)\left\(a+\frac{1}{3}\right\) \right\]-1 \)

\( \Longleftrightarrow a^2+b^2+c^2\ge 4(ab+bc+ca),\ c=-a-b \)

\( \Longleftrightarrow a^2+b^2+ab\ge0 \)

\( \Longleftrightarrow \left\(a+\frac{b}{2}\right\)^2+\frac{3b^2}{4}\ge0. \)

2. \( a\le1,\ b\le1 \Longrightarrow (a-1)(b-1)\ge 0 \)
\( \Longleftrightarrow a+b\le ab+1 \) si analoagele. Le adunam si tinem seama de \( ab+bc+ca=1 \):
\( \Longrightarrow a+b+c\le 2 \)
\( a^{2006}\le a \) si de aici concluzia.
Egalitatea are loc pentru \( a=b=1,\ c=0 \) si permutarile lor.

5)Avem ca \( \frac {1}{1 + x} = \frac {y}{1 + y} + \frac {z}{1 + z} \ge 2 \sqrt {\frac {yz}{(y + 1)(z + 1)}} \)

Scriind analoagele si inmultindu-le rezulta inegalitatea ceruta .