Trei inegalitati SHL-2010

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Andi Brojbeanu
Bernoulli
Posts: 294
Joined: Sun Mar 22, 2009 6:31 pm
Location: Targoviste (Dambovita)

Trei inegalitati SHL-2010

Post by Andi Brojbeanu »

1. Sa se arate ca, pentru orice numere reale \( x, y, z, t \)
\( \sqrt{(x^2+xy+y^2)(z^2+zt+t^2)}+\sqrt{(y^2-yz+z^2)(x^2-xt+t^2)}>(x+z)(y+t) \). Sa se precizeze in ce conditii se realizeaza egalitatea.
Calin Burdusel, Targoviste
2. Sa se arate ca, daca \( a, b, c \) sunt numere reale pozitive cu proprietatea \( abc=1 \), atunci
\( (1+a^3)(1+b^3)(1+c^3)\ge2(1+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}) \).
Dan Nedeianu, Drobeta Turnu Severin
3. Sa se arate ca numerele reale \( a,b \) satisfac relatia
\( (x+y+z)^3+a(x+y+z)(xy+yz+zx)+bxyz\ge 0, \forall x, y, z\ge 0 \) daca si numai daca exista \( r\ge 0, s\ge 0 \) astfel incat \( a=-4+r, b=9-9r+s \).
Dumitru Barac, Sibiu
Andi Brojbeanu
profesor, Liceul Teoretic "Lucian Blaga", Cluj-Napoca
Marius Mainea
Gauss
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Joined: Mon May 26, 2008 2:12 pm
Location: Gaesti (Dambovita)

Post by Marius Mainea »

1) \( LHS\ge (x+\frac{y}{2})(t+\frac{z}{2})+\frac{3yz}{4}+(y-\frac{z}{2})(x-\frac{t}{2})+\frac{3zt}{4}=RHS \)

2) Deconditionand obtinem :

\( \pro(x^3+y^3)\ge2x^2y^2z^2(xyz+\sum x^3) \)

adica \( \sum_{sym}x^6y^3\ge\sum_{sym}x^5y^2z^2 \) (Muirhead)
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