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Nguyễn Trọng Chiến · Accepted Answer

139:Đặt $x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\left(a,b,c>0\right)$GT $\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{3}{abc}\Rightarrow a+b+c=3$$\Rightarrow\dfrac{y^2}{xy^2+2x^2}=\dfrac{1}{b^2}:\left(\dfrac{1}{ab^2}+\dfrac{2}{a^2}\right)=\dfrac{1}{b^2}:\left(\dfrac{a+2b^2}{a^2b^2}\right)=\dfrac{a^2}{a+2b^2}=a-\dfrac{2ab^2}{a+2b^2}\ge a-\dfrac{2ab^2}{3b\sqrt[3]{ab}}=a-\dfrac{2}{3}\sqrt[3]{a^2b^2}\ge a-\dfrac{2}{9}\left(a+b+ab\right)$ Tương tự ta được: $\dfrac{x^2}{zx^2+2z^2}=\dfrac{c^2}{c+2a^2}=c-\dfrac{2ca^2}{c+2a^2}\ge c-\dfrac{2}{9}\left(c+a+ac\right)$$\dfrac{z^2}{yz^2+2y^2}=\dfrac{b^2}{b+2c^2}=b-\dfrac{2bc^2}{b+2c^2}\ge b-\dfrac{2}{9}\left(b+c+bc\right)$$\Rightarrow\dfrac{y^2}{xy^2+2x^2}+\dfrac{x^2}{zx^2+2z^2}+\dfrac{z^2}{yz^2+2z^2}\ge\left(a+b+c\right)-\dfrac{2}{9}\left(2a+2b+2c+ab+bc+ca\right)$ $\ge3-\dfrac{2}{9}\left[6+\dfrac{\left(a+b+c\right)^2}{3}\right]=3-\dfrac{2}{9}\left(6+\dfrac{9}{3}\right)=3-\dfrac{2}{9}\cdot9=1$Dấu bằng xảy ra $\Leftrightarrow a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=3$Câu trả lời: 139:Đặt $x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\left(a,b,c>0\right)$GT $\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{3}{abc}\Rightarrow a+b+c=3$$\Rightarrow\dfrac{y^2}{xy^2+2x^2}=\dfrac{1}{b^2}:\left(\dfrac{1}{ab^2}+\dfrac{2}{a^2}\right)=\dfrac{1}{b^2}:\left(\dfrac{a+2b^2}{a^2b^2}\right)=\dfrac{a^2}{a+2b^2}=a-\dfrac{2ab^2}{a+2b^2}\ge a-\dfrac{2ab^2}{3b\sqrt[3]{ab}}=a-\dfrac{2}{3}\sqrt[3]{a^2b^2}\ge a-\dfrac{2}{9}\left(a+b+ab\right)$ Tương tự ta được: $\dfrac{x^2}{zx^2+2z^2}=\dfrac{c^2}{c+2a^2}=c-\dfrac{2ca^2}{c+2a^2}\ge c-\dfrac{2}{9}\left(c+a+ac\right)$$\dfrac{z^2}{yz^2+2y^2}=\dfrac{b^2}{b+2c^2}=b-\dfrac{2bc^2}{b+2c^2}\ge b-\dfrac{2}{9}\left(b+c+bc\right)$$\Rightarrow\dfrac{y^2}{xy^2+2x^2}+\dfrac{x^2}{zx^2+2z^2}+\dfrac{z^2}{yz^2+2z^2}\ge\left(a+b+c\right)-\dfrac{2}{9}\left(2a+2b+2c+ab+bc+ca\right)$ $\ge3-\dfrac{2}{9}\left[6+\dfrac{\left(a+b+c\right)^2}{3}\right]=3-\dfrac{2}{9}\left(6+\dfrac{9}{3}\right)=3-\dfrac{2}{9}\cdot9=1$Dấu bằng xảy ra $\Leftrightarrow a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=3$

Khôi Nguyênx · Answer

câu trả lời :Đặt x=1a,y=1b,z=1c(a,b,c>0)x=1a,y=1b,z=1c(a,b,c>0)GT ⇒1ab+1bc+1ca=3abc⇒a+b+c=3⇒1ab+1bc+1ca=3abc⇒a+b+c=3⇒y2xy2+2x2=1b2:(1ab2+2a2)=1b2:(a+2b2a2b2)=a2a+2b2=a−2ab2a+2b2≥a−2ab23b3√ab=a−233√a2b2≥a−29(a+b+ab)Câu trả lời: câu trả lời :Đặt x=1a,y=1b,z=1c(a,b,c>0)x=1a,y=1b,z=1c(a,b,c>0)GT ⇒1ab+1bc+1ca=3abc⇒a+b+c=3⇒1ab+1bc+1ca=3abc⇒a+b+c=3⇒y2xy2+2x2=1b2:(1ab2+2a2)=1b2:(a+2b2a2b2)=a2a+2b2=a−2ab2a+2b2≥a−2ab23b3√ab=a−233√a2b2≥a−29(a+b+ab)

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