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Fie \( a,b,c \) trei numere reale pozitive astfel incat \( a+b+c=1 \). Sa se demonstreze ca :

\( \frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le \frac{3}{2}. \)

Inegalitatea este echivalenta cu \( \sum {\frac{a}{a+bc}}\le\frac{9}{4} \) sau \( \sum {\frac{a}{(a+b)(a+c)}\le \frac{9}{4}. \)

Notand a+b=z, b+c=x, c+a=y obtinem

\( \sum {\frac{x+y-z}{xy}}\le \frac{9}{x+y+z} \) care se reduce la inegalitatea lui Schur \( x^3+y^3+z^3+3xyz\ge x^2(y+z)+y^2(z+x)+z^2(x+y) \)

Luand \( a=\frac{x}{x+y+z},\ b=\frac{y}{x+y+z},\ c=\frac{z}{x+y+z} \) avem \( \sum\frac{a-bc}{a+bc}\le\frac{3}{2}\Longleftrightarrow\left(\sum\frac{a-bc}{a+bc}+1\right)\le\frac{3}{2}+3\Longleftrightarrow\sum\frac{bc}{a+bc}\ge\frac{3}{4}\ \left(*\right) \).

Dar \( \sum\frac{bc}{a+bc}=\sum\frac{\frac{yz}{\left(x+y++z\right)^{2}}}{\frac{x}{x+y+z}+\frac{yz}{\left(x+y+z\right)^{2}}}=\sum\frac{yz}{\left(x+y\right)\left(x+z\right)}=\frac{\sum yz\left(y+z\right)}{\prod\left(x+y\right)} \).

Acum \( \left(*\right)\Longleftrightarrow4\sum yz\left(y+z\right)\ge3\prod\left(x+y\right)\Longleftrightarrow\sum yz\left(y+z\right)\ge6xyz\Longleftrightarrow\sum\frac{y+z}{x}\ge6 \). Ultima inegalitate este evidenta, iar concluzia se impune.

Avem ca :

\( \frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\ge \frac{3}{2}\Leftrightarrow \left(\frac{a-bc}{a+bc}-1\right)+\left(\frac{b-ca}{b+ca}-1\right)+\left(\frac{c-ab}{c+ab}-1\right)\le -\frac{3}{2}\Leftrightarrow \frac{-2bc}{a+bc}+\frac{-2ca}{b+2ca}+\frac{-ab}{c+ab}\le -\frac{3}{2}\Leftrightarrow \frac{bc}{a+bc}+\frac{ca}{b+ca}+\frac{ab}{c+ab}\ge \frac{3}{4} \)

Din Cauchy-Schwartz avem ca :

\( \left(\frac{bc}{a+bc}+\frac{ca}{b+ca}+\frac{ab}{c+ab}\right)(3abc+a^2b^2+b^2c^2+c^2a^2)\ge (ab+bc+ca)^2\Leftrightarrow \frac{bc}{a+bc}+\frac{ca}{b+ca}+\frac{ab}{c+ab}\ge \frac{(ab+bc+ca)^2}{3abc+a^2b^2+b^2c^2+c^2a^2} \)

Deci ar fi de ajuns sa demonstram ca :

\(
\frac{(ab+bc+ca)^2}{3abc+a^2b^2+b^2c^2+c^2a^2}\ge \frac{3}{4} \)

\(
\Leftrightarrow 4(a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c))\ge 3(3abc+a^2b^2+b^2c^2+c^2a^2)\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc(a+b+c) \)

\( \Leftrightarrow 2(a^2b^2+b^2c^2+c^2a^2)\ge 2abc(a+b+c)\Leftrightarrow (ab-bc)^2+(bc-ca)^2+(ca-ab)^2\ge 0 \)