Numere pozitive; RUSIA-Con. Savin

mihai miculita
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Membru din: Mar Oct 26, 2010 9:21 pm
Localitate: ORADEA

Numere pozitive; RUSIA-Con. Savin

Mesaj de mihai miculita »

Aratati ca:
$\left\begin{array}{c}(b+c-a)^3+(a+c-b)^3+(a+b-c)^3=a^3+b^3+c^3\\ a;b;c\ge 0\end{array}\right\}$$\Rightarrow a=b=c.$
dangerous storm
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Membru din: Joi Iul 03, 2014 9:29 pm

Re: Numere pozitive; RUSIA-Con. Savin

Mesaj de dangerous storm »

Avem $(b+c-a)^3+(a+c-b)^3+(a+b-c)^3=a^3+b^3+c^3$ $\Longleftrightarrow$ $3a^2b+3a^2c+3ab^2$−$18abc+3ac^2+3b^2c+3bc^2=0$.Dar din AM-GM avem $a^2b+a^2c+ab^2+ac^2+b^2c+bc^2\ge 6abc$,deci $0=3a^2b+3a^2c+3ab^2$−$18abc+3ac^2+3b^2c+3bc^2$$=3(a^2b+a^2c+ab^2+ac^2+b^2c+bc^2-6abc)\ge 0$.
Asadar trebuie sa avem $a^2b+a^2c+ab^2+ac^2+b^2c+bc^2=6abc$,adica $a=b=c$.
mihai miculita
Mesaje: 1493
Membru din: Mar Oct 26, 2010 9:21 pm
Localitate: ORADEA

Re: Numere pozitive; RUSIA-Con. Savin

Mesaj de mihai miculita »

SOLUTIA MEA (detaliata):
Notand cu: $\left\{\begin{array}{c}b+c-a=x\\ a+c-b=y\\ a+b-c=z\end{array}\right\Rightarrow$$\left\{\begin{array}{c}2a=y+z\ge 0\\ 2b=x+z\ge 0\\ 2c=x+y\ge 0\end{array}\right ;$ avem:

$\left (b+c-a)^3+(a+c-b)^3+(a+b-c)^3=a^3+b^3+c^3\right|.8\Rightarrow$
$\Rightarrow 8.(b+c-a)^3+8.(a+c-b)^3+8.(a+b-c)^3=$$(2a)^3+(2b)^3+(2c)^3\Rightarrow$
$\Rightarrow 8x^3+8y^3+8z^3=(y+z)^3+(x+z)^3+(x+y)^3\Rightarrow$
$\Rightarrow 8x^3+8y^3+8z^3=(y^3+3y^2z+3yz^2+z^3)$$+(x^3+3x^2z+3z^2+z^3)+(x^3+3x^2y+3xy^2+y^3)\Rightarrow$
$\Rightarrow \left 6x^3+6y^3+6z^3=3y^2z+3yz^2+3x^2z+3z^2+3x^2y+3xy^2 \right|:3\Rightarrow$
$\Rightarrow 2x^3+2y^3+2z^3=y^2z+yz^2+x^2z+z^2+x^2y+xy^2\Rightarrow$
$\Rightarrow (x^3+y^3-x^2y-xy^2)+(x^3+z^3-x^2z-xz^2)$$+(y^3+z^3-y^2z-yz^2)=0\Rightarrow$
$\Rightarrow [(x+y)(x^2-xy+y^2)-xy(x+y)]+$$[(x+z)(x^2-xz+z^2)-xz(x+z)]+$
$+[(y+z)(y^2-yz+z^2)-yz(y+z)]=0\Rightarrow$
$\Rightarrow(x+y)(x^2-2xy+y^2)+(x+z)(x^2-2xz+z^2)$$+(y+z)(y^2-yz+z^2)=0\Rightarrow$
$\Rightarrow (x+y)(x-y)^2+(x+z)(x-z)^2+(y+z)(y-z)^2=0\Rightarrow$
$\Rightarrow x-y=x-z=y-z=0\Rightarrow x=y=z\Rightarrow$$b+c-a=a+c-b=a+b-c\Rightarrow \boxed{a=b=c}.$
Virgil Nicula
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Membru din: Sâm Oct 30, 2010 3:55 pm
Localitate: Bradenton, Florida

Re: Numere pozitive; RUSIA-Con. Savin

Mesaj de Virgil Nicula »

Folosim identitatea (x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x) si inegalitatea (x+y)(y+z)(z+x)\ge 8xyz (cu egalitate pentru x=y=z), unde x,y,z

sunt numere reale nenegative in situatia particulara x:= b+c-a , y:=c+a-b si z:=a+b-c, unde x+y+z=a+b+c si y+z=2a, z+x=2b, x+y=2c. Prin urmare,

(a+b+c)^3=(b+c-a)^3+(c+a-b)^3+(a+b-c)^3+24abc si abc\ge (b+c-a)(c+a-b)(a+b-c). In comcluzie, (b+c-a)^3+(c+a-b)^3+(a+b-c)^3=a^3+b^3+c^3

<=> (a+b+c)^3-24abc=a^3+b^3+c^3 <=> 3(a+b)(b+c)(c+a)=24abc <=> (a+b)(b+c)(c+a)=8abc <=> a=b=c
.

Observatie. Vad ca la altii merge LaTeX - ul ... Poate imi spune cineva ce trebuie sa fac ca LaTeX - ul sa mearga si la mine. Multumesc.
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