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Đặt A = y 2 + y z + z 2 + z 2 + z x + x 2 + x 2 + x y + y 2 = ( y + z 2 ) 2 + ( 3 z 2 ) 2 + ( z + x 2 ) 2 + ( 3 x 2 ) 2 + ( x + y 2 ) 2 + ( 3 y 2 ) 2 A=\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}+\sqrt{x^2+xy+y^2}\\=\sqrt{\left ( y+\frac{z}{2} \right )^2+\left ( \frac{\sqrt{3}z}{2} \right )^2}+\sqrt{\left ( z+\frac{x}{2} \right )^2+\left ( \frac{\sqrt{3}x}{2} \right )^2}+\sqrt{\left ( x+\frac{y}{2} \right )^2+\left ( \frac{\sqrt{3}y}{2} \right )^2} A = y 2 + y z + z 2 + z 2 + z x + x 2 + x 2 + x y + y 2 = ( y + 2 z ) 2 + ( 2 3 z ) 2 + ( z + 2 x ) 2 + ( 2 3 x ) 2 + ( x + 2 y ) 2 + ( 2 3 y ) 2
Áp dụng BĐT Minkovsky ta có:
A ≥ ( x + y + z + x + y + z 2 ) 2 + ( 3 x 2 + 3 y 2 + 3 z 2 ) 2 = 3 ( x + y + z ) 2 = 3 ( x + y + z ) A\geq \sqrt{\left ( x+y+z+\frac{x+y+z}{2} \right )^2+\left ( \frac{\sqrt{3}x}{2}+\frac{\sqrt{3}y}{2}+\frac{\sqrt{3}z}{2} \right )^2}=\sqrt{3(x+y+z)^2}=\sqrt{3}(x+y+z) A ≥ ( x + y + z + 2 x + y + z ) 2 + ( 2 3 x + 2 3 y + 2 3 z ) 2 = 3 ( x + y + z ) 2 = 3 ( x + y + z )
Suy ra x y + y z + z x A ≤ x y + y z + z x 3 ( x + y + z ) ≤ ( x + y + z ) 2 3 3 ( x + y + z ) = x + y + z 3 3 ( d p c m ) \frac{xy+yz+zx}{A}\leq \frac{xy+yz+zx}{\sqrt{3}(x+y+z)}\leq \frac{(x+y+z)^2}{3\sqrt{3}(x+y+z)}=\frac{x+y+z}{3\sqrt{3}}(dpcm) A x y + y z + z x ≤ 3 ( x + y + z ) x y + y z + z x ≤ 3 3 ( x + y + z ) ( x + y + z ) 2 = 3 3 x + y + z ( d p c m )
Dấu = xảy ra khi x = y = z x=y=z x = y = z