Fie \( a,b,c \in \mathb{R}_+^* \), astfel incat \( (a+b)^2+(b+c)^2+(c+a)^2+(a+b)(b+c)(c+a)=4. \) Sa se arate ca \( \frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\geq \frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}. \)
Andrei Vrajitoarea, Recreatii Matematice 1/2009
Inegalitate conditionata
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Fie \( x=\frac{b+c}{2} \) ,\( y=\frac{c+a}{2} \) , \( z =\frac{a+b}{2} \) ,
Din ipoteza rezulta ca \( x\le 1 \) , \( y\le 1 \) ,\( z\le 1 \)
atunci \( a=y+z-x \) , \( b=z+x-y \) , \( c=x+y-z \) si inegalitatea de aratat devine
\( \sum a^2\ge\sum ab(a+b) \) sau
\( \sum (y+z-x)^2\ge \sum (z+y-x)(z+x-y)2z \) sau prin desfaceera parantezelor
\( 3\sum x^2+2\sum xy\ge 2\sum x^3-2\sum x(y^2+z^2)+12xyz \) care este adevarata (demonstrati !)
Din ipoteza rezulta ca \( x\le 1 \) , \( y\le 1 \) ,\( z\le 1 \)
atunci \( a=y+z-x \) , \( b=z+x-y \) , \( c=x+y-z \) si inegalitatea de aratat devine
\( \sum a^2\ge\sum ab(a+b) \) sau
\( \sum (y+z-x)^2\ge \sum (z+y-x)(z+x-y)2z \) sau prin desfaceera parantezelor
\( 3\sum x^2+2\sum xy\ge 2\sum x^3-2\sum x(y^2+z^2)+12xyz \) care este adevarata (demonstrati !)