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Daca \( a,b,c\in(0,\infty) \) cu \( a+b+c=1 \) atunci

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

MARIUS MAINEA wrote:Daca \( a,b,c\in(0,\infty) \) cu \( a+b+c=1 \) atunci \( \frac{a^2+a}{a+bc}+\frac{b^2+b}{b+ca}+\frac{c^2+c}{c+ab}=3 \)

Draguta ! Se reduce la identitatea evidenta

\( \sum a^2(b+c)=\sum bc(b+c) \) , adica \( \sum \frac {a^2-bc}{(a+b)(a+c)}=0 \) .

Intr-adevar, \( \sum\left(\frac {a^2+a}{a+bc}-1\right)=\sum \frac {a^2-bc}{(a+b)(a+c)}=0\ \) \( \Longrightarrow\ \) \( \sum\frac {a^2+a}{a+bc}=3 \) .

Cu substitutiile \( a=\frac{x}{x+y+z},b=\frac{y}{x+y+z},c=\frac{z}{x+y+z} \) unde \( x,y,z>0 \) avem:\( \sum\frac{a^{2}+a}{a+bc}=\sum\frac{\frac{x^{2}}{\left(x+y+z\right)^{2}}+\frac{x}{x+y+z}}{\frac{yz}{\left(x+y+z\right)^{2}}+\frac{x}{x+y+z}}=\sum\frac{x\left(x+y+x+z\right)}{yz+x\left(x+y+z\right)}=\sum\frac{x\left(x+y\right)+x\left(x+z\right)}{\left(x+y\right)\left(x+z\right)}=\sum\frac{x}{x+y}+\sum\frac{x}{z+x}=3 \), ceea ce trebuia aratat.