Bài 3. Từ giả thiết $a+b+c=3abc\Rightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=3$
Đặt $\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\Rightarrow xy+yz+zx=3$ và x, y, z > 0.
Khi đó $P=\frac{x}{x+1+3yz}+\frac{y}{y+1+3zx}+\frac{z}{z+1+3xy}$
$P=\frac{x^{2}}{x^{2}+x+3xyz}+\frac{y^{2}}{y^{2}+y+3xyz}+\frac{z^{2}}{z^{2}+z+3xyz}$
$\geq \frac{\left ( x+y+z \right )^{2}}{9xyz+x^{2}+y^{2}+z^{2}+x+y+z}$
Áp dụng Cauchy, ta có $\left ( x+y+z \right )\left ( xy+yz+zx \right )\geq \sqrt[3]{xyz}.\sqrt[3]{x^{2}y^{2}z^{2}}=9xyz$
$\Rightarrow 9xyz\leq 3\left ( x+y+z \right )\Leftrightarrow 3xyz\leq x+y+z$
Lại có $xy+yz+zx\geq 3\sqrt[3]{x^{2}y^{2}z^{2}}\Rightarrow xyz\leq 1$
Suy ra $9xyz=3xyz+6xyz\leq x+y+z+2\left ( xy+yz+zx \right )$
$\Rightarrow 9xyz+x^{2}+y^{2}+z^{2}+x+y+z\leq \left ( x+y+z \right )^{2}+2\left ( x+y+z \right )$
Lại có $\left ( x+y+z \right )^{2}\geq 3\left ( xy+yz+zx \right )=9\Rightarrow x+y+z\geq 3$
Vậy $P\geq 1-\frac{2}{3+2}=\frac{3}{5}$