Cho các số dương a, b, c tm a+b+c=abc.
Tìm GTNN của A= a/(c căn (a^2+1)) + b/(a căn(b^2+1)) + c/(b(căn(c^2+1))
Các bạn giúp mk với! Thanks
[tex]\frac{a}{c.\sqrt{a^2+1}}=\frac{1}{c.\sqrt{1+\frac{1}{a^2}}}\\\\ +, a+b+c=abc\\\\ => \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\\\\ +, (\frac{1}{a};\frac{1}{b};\frac{1}{c})=(x;y;z)\\\\ =>+, xy+yz+zx=1\\\\ +, A=\frac{z}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+y^2}}+\frac{y}{\sqrt{1+z^2}}\\\\ =\frac{z}{\sqrt{(x+y).(x+z)}}+\frac{x}{\sqrt{(x+y).(y+z)}}+\frac{y}{\sqrt{(y+z).(x+z)}}\\\\ \geq \frac{2z}{2x+y+z}+\frac{2x}{x+2y+z}+\frac{2y}{x+y+2z}\\\\ => \frac{A}{2}\geq \frac{z^2}{2xz+yz+z^2}+\frac{x^2}{x^2+2xy+xz}+\frac{y^2}{xy+y^2+2yz}=B\\\\ => B\geq \frac{(x+y+z)^2}{x^2+y^2+z^2+2xy+2yz+2zx+xy+yz+xz}\\\\ \geq \frac{(x+y+z)^2}{(x+y+z)^2+\frac{(x+y+z)^2}{3}}=\frac{1}{1+\frac{1}{3}}=\frac{3}{4}\\\\ => A\geq \frac{3}{2}[/tex]
dấu "=" <=> a=b=c= căn 3
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