Lời giải:
Ta có:
\(P=\frac{y-2}{x^2}+\frac{z-2}{y^2}+\frac{x-2}{z^2}\)
\(P=\frac{(x-1)+(y-1)}{x^2}-\frac{1}{x}+\frac{(y-1)+(z-1)}{y^2}-\frac{1}{y}+\frac{(x-1)+(z-1)}{z^2}-\frac{1}{z}\)
\(P=(x-1)\left(\frac{1}{x^2}+\frac{1}{z^2}\right)+(y-1)\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+(z-1)\left(\frac{1}{y^2}+\frac{1}{z^2}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Áp dụng BĐT AM-GM:
\(\frac{1}{x^2}+\frac{1}{z^2}\geq \frac{2}{xz}; \frac{1}{x^2}+\frac{1}{y^2}\geq \frac{2}{xy}; \frac{1}{y^2}+\frac{1}{z^2}\geq \frac{2}{yz}\)
Kết hợp với \(x-1,y-1,z-1>0\) theo đkđb thì:
\(P\geq \frac{2(x-1)}{xz}+\frac{2(y-1)}{xy}+\frac{2(z-1)}{yz}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\Leftrightarrow P\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}-2\left(\frac{1}{xz}+\frac{1}{xy}+\frac{1}{yz}\right)\)
\(\Leftrightarrow P\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{2(x+y+z)}{xyz}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-2\)
\(\Leftrightarrow P\geq \frac{xy+yz+xz}{xyz}-2(*)\)
Ta có một hệ quả quen thuộc của BĐT AM-GM:
\((xy+yz+xz)^2\geq 3xyz(x+y+z)\)
Mà \(x+y+z=xyz\Rightarrow (xy+yz+xz)^2\geq 3x^2y^2z^2\)
\(\Rightarrow \frac{xy+yz+xz}{xyz}\geq \sqrt{3}(**)\)
Từ \((*); (**)\Rightarrow P\geq \sqrt{3}-2\)
Dấu bằng xảy ra khi \(x=y=z=\sqrt{3}\)
Lời giải:
Ta có:
P=y−2x2+z−2y2+x−2z2P=y−2x2+z−2y2+x−2z2
P=(x−1)+(y−1)x2−1x+(y−1)+(z−1)y2−1y+(x−1)+(z−1)z2−1zP=(x−1)+(y−1)x2−1x+(y−1)+(z−1)y2−1y+(x−1)+(z−1)z2−1z
P=(x−1)(1x2+1z2)+(y−1)(1x2+1y2)+(z−1)(1y2+1z2)−(1x+1y+1z)P=(x−1)(1x2+1z2)+(y−1)(1x2+1y2)+(z−1)(1y2+1z2)−(1x+1y+1z)
Áp dụng BĐT AM-GM:
1x2+1z2≥2xz;1x2+1y2≥2xy;1y2+1z2≥2yz1x2+1z2≥2xz;1x2+1y2≥2xy;1y2+1z2≥2yz
Kết hợp với x−1,y−1,z−1>0x−1,y−1,z−1>0 theo đkđb thì:
P≥2(x−1)xz+2(y−1)xy+2(z−1)yz−(1x+1y+1z)P≥2(x−1)xz+2(y−1)xy+2(z−1)yz−(1x+1y+1z)
⇔P≥1x+1y+1z−2(1xz+1xy+1yz)⇔P≥1x+1y+1z−2(1xz+1xy+1yz)
⇔P≥1x+1y+1z−2(x+y+z)xyz=1x+1y+1z−2⇔P≥1x+1y+1z−2(x+y+z)xyz=1x+1y+1z−2
⇔P≥xy+yz+xzxyz−2(∗)⇔P≥xy+yz+xzxyz−2(∗)
Ta có một hệ quả quen thuộc của BĐT AM-GM:
(xy+yz+xz)2≥3xyz(x+y+z)(xy+yz+xz)2≥3xyz(x+y+z)
Mà x+y+z=xyz⇒(xy+yz+xz)2≥3x2y2z2x+y+z=xyz⇒(xy+yz+xz)2≥3x2y2z2
⇒xy+yz+xzxyz≥√3(∗∗)⇒xy+yz+xzxyz≥3(∗∗)
Từ (∗);(∗∗)⇒P≥√3−2(∗);(∗∗)⇒P≥3−2
Dấu bằng xảy ra khi x=y=z=√3
Ta có:
P=y−2x2+z−2y2+x−2z2P=y−2x2+z−2y2+x−2z2
P=(x−1)+(y−1)x2−1x+(y−1)+(z−1)y2−1y+(x−1)+(z−1)z2−1zP=(x−1)+(y−1)x2−1x+(y−1)+(z−1)y2−1y+(x−1)+(z−1)z2−1z
P=(x−1)(1x2+1z2)+(y−1)(1x2+1y2)+(z−1)(1y2+1z2)−(1x+1y+1z)P=(x−1)(1x2+1z2)+(y−1)(1x2+1y2)+(z−1)(1y2+1z2)−(1x+1y+1z)
Áp dụng BĐT AM-GM:
1x2+1z2≥2xz;1x2+1y2≥2xy;1y2+1z2≥2yz1x2+1z2≥2xz;1x2+1y2≥2xy;1y2+1z2≥2yz
Kết hợp với x−1,y−1,z−1>0x−1,y−1,z−1>0 theo đkđb thì:
P≥2(x−1)xz+2(y−1)xy+2(z−1)yz−(1x+1y+1z)P≥2(x−1)xz+2(y−1)xy+2(z−1)yz−(1x+1y+1z)
⇔P≥1x+1y+1z−2(1xz+1xy+1yz)⇔P≥1x+1y+1z−2(1xz+1xy+1yz)
⇔P≥1x+1y+1z−2(x+y+z)xyz=1x+1y+1z−2⇔P≥1x+1y+1z−2(x+y+z)xyz=1x+1y+1z−2
⇔P≥xy+yz+xzxyz−2(∗)⇔P≥xy+yz+xzxyz−2(∗)
Ta có một hệ quả quen thuộc của BĐT AM-GM:
(xy+yz+xz)2≥3xyz(x+y+z)(xy+yz+xz)2≥3xyz(x+y+z)
Mà x+y+z=xyz⇒(xy+yz+xz)2≥3x2y2z2x+y+z=xyz⇒(xy+yz+xz)2≥3x2y2z2
⇒xy+yz+xzxyz≥√3(∗∗)⇒xy+yz+xzxyz≥3(∗∗)
Từ (∗);(∗∗)⇒P≥√3−2(∗);(∗∗)⇒P≥3−2
Dấu bằng xảy ra khi x=y=z=√3
