$\Leftrightarrow$ a2b2+b2c2+c2a2≥3(a4b2c2+a2b4c2+a2b2c4)" role="presentation" style="font-size: 15px; display: inline; position: relative;">a2b2+b2c2+c2a2≥3(a4b2c2+a2b4c2+a2b2c4)−−−−−−−−−−−−−−−−−−−−−−√a2b2+b2c2+c2a2≥3(a4b2c2+a2b4c2+a2b2c4)Đặt: x=a2b2,y=b2c2,z=c2a2" role="presentation" style="font-size: 15px; display: inline; position: relative;">x=a2b2,y=b2c2,z=c2a2x=a2b2,y=b2c2,z=c2a2.Ta cm: x+y+z≥√3(xy+yz+zx)⇔(x+y+z)2≥3(xy+yz+zx) , luôn đúng ⇒ đpcmĐẳng thức xảy ra ⇔ x=y=z⇔a=b=c
$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq \sqrt{3(a^{2}+b^{2}+c^{2})}$$\Leftrightarrow a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\geq \sqrt{3(a^{4}b^{2}c^{2}+a^{2}b^{4}c^{2}+a^{2}b^{2}c^{4})}$Đặt: $x=a^{2}b^{2}; y=b^{2}c^{2}; z=c^{2}a^{2} (x, y, z >0)$Ta cm: $x+y+z\geq \sqrt{3(xy+yz+zx)} $ ⇔(x+y+z)2≥3(xy+yz+zx) , luôn đúng ⇒ đpcmĐẳng thức xảy ra ⇔ x=y=z⇔a=b=c