\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\)
\(5x=2z\Rightarrow\frac{x}{2}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow x=2k;y=3k;z=5k\)
\(\Rightarrow\left(2k\right)^3+\left(3k\right)^3-2k\cdot3k\cdot5k=40\)
\(\Rightarrow k^3\cdot8+k^3\cdot27-k^3\cdot30=40\)
\(\Rightarrow k^3\left(8+27-30\right)=40\)
\(\Rightarrow k^3=8\)
\(\Rightarrow k=2\)
\(\Rightarrow\hept{\begin{cases}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot5=10\end{cases}}\)
Có xy+yz+zx=xyzxy+yz+zx=xyz⇔⇔xy+yz+zxxyz=1xy+yz+zxxyz=1⇔⇔1x+1y+1z=11x+1y+1z=1
x2yy+2x+y2zz+2y+z2xx+2z=11x2+2xy+11y2+2yz+11z2+2zx≥91x2+1y2+1z2+2(1xy+1yz+1zx)x2yy+2x+y2zz+2y+z2xx+2z=11x2+2xy+11y2+2yz+11z2+2zx≥91x2+1y2+1z2+2(1xy+1yz+1zx)
=9(1x+1y+1z)2=912=9=9(1x+1y+1z)2=912=9
Dấu "=" ko xảy ra ⇒⇒x2yy+2x+y2zz+2y+z2xx+2z>9
