Toán 9 Cho x,y,z>0x,y,z>0. Chứng minh x33xyx2+xy+y2\frac{\sum x}{3\sqrt{3}}\geq \frac{\sum xy}{\sum \sqrt{x^2+xy+y^2}}

Ann Lee

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Đặt A=y2+yz+z2+z2+zx+x2+x2+xy+y2=(y+z2)2+(3z2)2+(z+x2)2+(3x2)2+(x+y2)2+(3y2)2A=\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}
Áp dụng BĐT Minkovsky ta có:
A(x+y+z+x+y+z2)2+(3x2+3y2+3z2)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)
Suy ra xy+yz+zxAxy+yz+zx3(x+y+z)(x+y+z)233(x+y+z)=x+y+z33(dpcm)\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)
Dấu = xảy ra khi x=y=zx=y=z
 
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