1. Vì x, y, z > 0
\(xy+yz+zx\ge2xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge2\)
Suy ra:
\(\dfrac{1}{x}\ge1-\dfrac{1}{y}+1-\dfrac{1}{z}=\dfrac{y-1}{y}+\dfrac{z-1}{z}\ge2\sqrt{\dfrac{\left(y-1\right)\left(z-1\right)}{yz}}\). (1)
Tương tự \(\dfrac{1}{y}\ge2\sqrt{\dfrac{\left(z-1\right)\left(x-1\right)}{zx}}\) (2)
và \(\dfrac{1}{z}\ge2\sqrt{\dfrac{\left(x-1\right)\left(y-1\right)}{xy}}\) (3)
Nhân (1), (2), (3) với nhau theo vế ta được
\(\dfrac{1}{xyz}\ge\dfrac{8\left(x-1\right)\left(y-1\right)\left(z-1\right)}{xyz}\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)\le\dfrac{1}{8}\)
Đẳng thức xảy ra \(\Leftrightarrow x=y=z=\dfrac{3}{2}\)
\(\dfrac{c+1}{c+3}\ge\dfrac{1}{a+2}+\dfrac{3}{b+4}\)
\(\Leftrightarrow1-\dfrac{2}{c+3}\ge\dfrac{1}{a+2}+\dfrac{3}{b+4}\)
\(\Leftrightarrow1-\dfrac{1}{a+2}\ge\dfrac{3}{b+4}+\dfrac{2}{c+3}\ge2\sqrt{\dfrac{6}{\left(b+4\right)\left(c+3\right)}}\)
Hay \(\dfrac{a+1}{a+2}\ge2\sqrt{\dfrac{6}{\left(b+4\right)\left(c+3\right)}}\) (1)
Tương tự \(\dfrac{b+1}{b+4}\ge2\sqrt{\dfrac{2}{\left(c+3\right)\left(a+2\right)}}\) (2)
và \(\dfrac{c+1}{c+3}\ge2\sqrt{\dfrac{3}{\left(a+2\right)\left(b+4\right)}}\) (3)
Nhân (1), (2), (3) vế theo vế
\(\dfrac{\left(a+1\right)\left(b+1\right)\left(c+1\right)}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\ge8.\dfrac{6}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\)
\(\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge48\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=5\\c=3\end{matrix}\right.\)