Question

Cho a b c là các số thực dương thỏa mãn \(\left\{{}\begin{matrix}a+b+c=5\\\sqrt{a}+\sqrt{b}+\sqrt{c}=3\end{matrix}\right.\)

CMR :\(\frac{\sqrt{a}}{a+2}+\frac{\sqrt{b}}{b+2}+\frac{\sqrt{c}}{c+2}=\frac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)

Answer 1

Cho dễ nhìn thì \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\)

\(x+y+z=3\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=9\)

\(\Rightarrow xy+yz+zx=2\)

\(VT=\sum\frac{x}{x^2+2}=\sum\frac{x}{x^2+xy+yz+zx}=\sum\frac{x}{\left(x+y\right)\left(x+z\right)}\)

\(=\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{4}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(VP=\frac{4}{\sqrt{\left(x+y\right)\left(x+z\right)\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(z+x\right)}}=\frac{4}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=VT\) (đpcm)