Inegalitate in 4 variabile!

[maxim bogdan]

Fie \( a,b,c,d>0 \), cu proprietatea ca: \( a^2+b^2+c^2+d^2=1. \) Demonstrati inegalitatea:

\( (1-a)(1-b)(1-c)(1-d)\geq abcd \)

[Marius Mainea]

Se arata ca \( (1-a)(1-b)\ge cd \) si

\( (1-c)(1-d)\ge ab \)

[maxim bogdan]

Poti sa detaliezi solutia, te rog.

[Marius Mainea]

\( cd\le\frac{c^2+d^2}{2}=\frac{1-a^2-b^2}{2}\le(1-a)(1-b) \) \( \Longleftrightarrow (1-a-b)^2\ge 0 \)

Analog \( ab\le (1-c)(1-d) \) si prin inmultire obtinem concluzia.

[maxim bogdan]

Solutie. Presupunem ca: \( \frac{(1-a)(1-b)(1-c)(1-d)}{abcd}=p^4, p<1. \) Facem substitutia: \( x=\frac{1-a}{pa};y=\frac{1-b}{pb};z=\frac{1-c}{pc};t=\frac{1-d}{pd}. \)

\( x=\frac{1-a}{pa}\Rightarrow px=\frac{1-a}{a}\Rightarrow 1+px=\frac{1}{a}\Rightarrow \frac{1}{(1+px)^2}=a^2. \) Analog: \( \frac{1}{(1+py)^2}=b^2;\frac{1}{(1+pz)^2}=c^2;\frac{1}{(1+pt)^2}=d^2. \)

Conditia din enunt este echivalenta cu: \( \sum_{cyc}\frac{1}{(1+px)^2}=1 \), unde \( xyzt=1 \). Cum \( p<1 \):

\( \Rightarrow1=\sum_{cyc}\frac{1}{(1+px)^2}>\sum_{cyc}\frac{1}{(1+x)^2}. \)

Problema se reduce la a demonstra ca:

\( \frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+t)^2}\geq 1. \) (inegalitate data la China Team Selection Test 2005).

Se foloseste cunoscuta lema: Fie \( x,y\in\mathbb{R_{+}} \). Atunci are loc urmatoarea inegalitate:

\( \frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}\geq \frac{1}{1+xy}. \)

Dupa evectuarea calculelor inegalitatea e echivalenta cu: \( xy(x-y)^2+(1-xy)^2\geq 0 \) care este evidenta. Mai exista o demonstratie a acestei leme utilizand inegalitatea Cauchy-Buniakowski-Schwarz apartinand lui Darij Grinberg. (vezi aici.)

Aplicand lema avem: \( (\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2})+(\frac{1}{(1+z)^2}+\frac{1}{(1+t)^2})\geq \frac{1}{1+xy}+\frac{1}{1+zt}=\frac{1+xy+zt+xyzt}{(1+xy)(1+zt)}=1. \)

Deci presupunerea facuta este falsa si concluzia este evidenta!

[alex2008]

O solutie asemanatoare cu cea a domnului Mainea :

Demonstram ca : \( (1-a)(1-b)(a^2+b^2)\ge ab(c^2+d^2) \)

si \( (1-c)(1-d)(c^2+d^2)\ge cd(a^2+b^2) \)

\( 2(1-a)(1-b)(a^2+b^2)\ge 2ab(c^2+d^2)\Leftrightarrow ((a+b-1)^2+c^2+d^2)(a^2+b^2)-2ab(c^2+d^2)\ge 0\Leftrightarrow (a+b-1)^2(a^2+b^2)+(a-b)^2(c^2+d^2)\ge 0 \) , unde am folosit identitatea \( 2(1-a)(1-b)=(a+b-1)^2+c^2+d^2 . \)

Prin inmultire rezulta concluzia .