Daca a,b,c sunt pozitive si a+b+c=1 , atunci
\( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\le \frac{1-a+b}{2c}+\frac{1-b+c}{2a}+\frac{1-c+a}{2b} \)
R.M.T.
Inegalitate rationala
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Marius Mainea
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Inlocuind pe \( 1 \) cu \( a+b+c \), inegalitatea devine:
\( \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}\leq\frac{a+b+c-a+b}{2c}+\frac{a+b+c-b+c}{2a}+\frac{a+b+c-c+a}{2b} \) sau \( \sum{\frac{a}{b+c}}+3\leq \sum{\frac{2a+b}{2b}} \).
\( RHS=\sum {\frac{2a+b}{2b}}=\sum{\frac{2a}{2b}}+\sum{\frac{b}{2b}}=\sum{\frac{a}{b}}+\sum{\frac{1}{2}}=\sum{\frac{a}{b}}+\frac{3}{2}. \).
Deci, inegalitatea se mai scrie astfel: \( \sum{\frac{a}{b+c}}+3\leq\sum{\frac{a}{b}}+\frac{3}{2} \) sau \( \sum{\frac{a}{b}}-\sum{\frac{a}{b+c}}\geq\frac{3}{2} \).
\( LHS=\sum{\frac{a}{b}}-\sum{\frac{a}{b+c}}=\sum{\frac{a(b+c)-ab}{b(b+c)}}=\sum{\frac{ab+ac-ab}{b(b+c)}}=\sum{\frac{ac}{b(b+c)}}=\sum{\frac{(ac)^2}{abc(b+c)}} \).
Dar din inegalitatea Cauchy-Buniakovski-Schwarz, avem: \( LHS=\sum{\frac{(ac)^2}{abc(b+c)}}\geq \frac{(ab+bc+ac)^2}{2abc(a+b+c)} \)
Facand notatiile \( ab=x, bc=y, ca=z \), obtinem ca \( LHS\geq \frac{(x+y+z)^2}{2(xy+yz+zx)}=\frac{x^2+y^2+z^2}{2(xy+yz+zx)}+\frac{2(xy+yz+zx)}{2(xy+yz+zx)}=\frac{x^2+y^2+z^2}{2(xy+yz+zx)}+1 \).
Din inegalitatea cunoscuta \( x^2+y^2+z^2\geq xy+yz+zx \), cu egalitate pentru \( x=y=z \), obtinem ca \( LHS\geq\frac{x^2+y^2+z^2}{2(xy+yz+zx)}+1\geq \frac{1}{2}+1=\frac{3}{2}=RHS \), cu egalitate pentru \( a=b=c=\frac{1}{3} \).
\( \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}\leq\frac{a+b+c-a+b}{2c}+\frac{a+b+c-b+c}{2a}+\frac{a+b+c-c+a}{2b} \) sau \( \sum{\frac{a}{b+c}}+3\leq \sum{\frac{2a+b}{2b}} \).
\( RHS=\sum {\frac{2a+b}{2b}}=\sum{\frac{2a}{2b}}+\sum{\frac{b}{2b}}=\sum{\frac{a}{b}}+\sum{\frac{1}{2}}=\sum{\frac{a}{b}}+\frac{3}{2}. \).
Deci, inegalitatea se mai scrie astfel: \( \sum{\frac{a}{b+c}}+3\leq\sum{\frac{a}{b}}+\frac{3}{2} \) sau \( \sum{\frac{a}{b}}-\sum{\frac{a}{b+c}}\geq\frac{3}{2} \).
\( LHS=\sum{\frac{a}{b}}-\sum{\frac{a}{b+c}}=\sum{\frac{a(b+c)-ab}{b(b+c)}}=\sum{\frac{ab+ac-ab}{b(b+c)}}=\sum{\frac{ac}{b(b+c)}}=\sum{\frac{(ac)^2}{abc(b+c)}} \).
Dar din inegalitatea Cauchy-Buniakovski-Schwarz, avem: \( LHS=\sum{\frac{(ac)^2}{abc(b+c)}}\geq \frac{(ab+bc+ac)^2}{2abc(a+b+c)} \)
Facand notatiile \( ab=x, bc=y, ca=z \), obtinem ca \( LHS\geq \frac{(x+y+z)^2}{2(xy+yz+zx)}=\frac{x^2+y^2+z^2}{2(xy+yz+zx)}+\frac{2(xy+yz+zx)}{2(xy+yz+zx)}=\frac{x^2+y^2+z^2}{2(xy+yz+zx)}+1 \).
Din inegalitatea cunoscuta \( x^2+y^2+z^2\geq xy+yz+zx \), cu egalitate pentru \( x=y=z \), obtinem ca \( LHS\geq\frac{x^2+y^2+z^2}{2(xy+yz+zx)}+1\geq \frac{1}{2}+1=\frac{3}{2}=RHS \), cu egalitate pentru \( a=b=c=\frac{1}{3} \).