Câu hỏi của Quoc Tran Anh Le

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Hồng Phúc · Accepted Answer

C280:Áp dụng BĐT AM-GM và BĐT BSC:$\dfrac{1}{\sqrt{x+3y}}+\sqrt{x+3y}\ge2\Rightarrow\dfrac{1}{\sqrt{x+3y}}\ge2-\sqrt{x+3y}$$\dfrac{1}{\sqrt{y+3z}}+\sqrt{y+3z}\ge2\Rightarrow\dfrac{1}{\sqrt{y+3z}}\ge2-\sqrt{y+3z}$$\dfrac{1}{\sqrt{z+3x}}+\sqrt{z+3x}\ge2\Rightarrow\dfrac{1}{\sqrt{z+3x}}\ge2-\sqrt{z+3x}$$\Rightarrow P=\dfrac{1}{\sqrt{x+3y}}+\dfrac{1}{\sqrt{y+3z}}+\dfrac{1}{\sqrt{z+3x}}$$\ge6-\left(\sqrt{x+3y}+\sqrt{y+3z}+\sqrt{z+3x}\right)$$\ge6-\sqrt{3\left(x+3y+y+3z+z+3x\right)}$$=6-\sqrt{12\left(x+y+z\right)}=3$$minP=3\Leftrightarrow a=b=c=\dfrac{1}{4}$Câu trả lời: C280:Áp dụng BĐT AM-GM và BĐT BSC:$\dfrac{1}{\sqrt{x+3y}}+\sqrt{x+3y}\ge2\Rightarrow\dfrac{1}{\sqrt{x+3y}}\ge2-\sqrt{x+3y}$$\dfrac{1}{\sqrt{y+3z}}+\sqrt{y+3z}\ge2\Rightarrow\dfrac{1}{\sqrt{y+3z}}\ge2-\sqrt{y+3z}$$\dfrac{1}{\sqrt{z+3x}}+\sqrt{z+3x}\ge2\Rightarrow\dfrac{1}{\sqrt{z+3x}}\ge2-\sqrt{z+3x}$$\Rightarrow P=\dfrac{1}{\sqrt{x+3y}}+\dfrac{1}{\sqrt{y+3z}}+\dfrac{1}{\sqrt{z+3x}}$$\ge6-\left(\sqrt{x+3y}+\sqrt{y+3z}+\sqrt{z+3x}\right)$$\ge6-\sqrt{3\left(x+3y+y+3z+z+3x\right)}$$=6-\sqrt{12\left(x+y+z\right)}=3$$minP=3\Leftrightarrow a=b=c=\dfrac{1}{4}$

Mashiro Shiina · Answer

Bài 7) $bđt\Leftrightarrow4\left(a+b+c\right)\left(a^2+b^2+c^2\right)-3\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3$$\Leftrightarrow a^3+b^3+c^3+4ab\left(a+b\right)+4bc\left(b+c\right)+4ac\left(a+c\right)\ge\left(a+b+c\right)^3$$\Leftrightarrow4ab\left(a+b\right)+4bc\left(b+c\right)+4ac\left(a+c\right)\ge3ab\left(a+b\right)+3bc\left(b+c\right)+3ac\left(a+c\right)+6abc$$\Leftrightarrow ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\ge6abc$$\Leftrightarrow\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6$(Đúng theo Cô Si)"=" khi a=b=c=1Câu trả lời: Bài 7) $bđt\Leftrightarrow4\left(a+b+c\right)\left(a^2+b^2+c^2\right)-3\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3$$\Leftrightarrow a^3+b^3+c^3+4ab\left(a+b\right)+4bc\left(b+c\right)+4ac\left(a+c\right)\ge\left(a+b+c\right)^3$$\Leftrightarrow4ab\left(a+b\right)+4bc\left(b+c\right)+4ac\left(a+c\right)\ge3ab\left(a+b\right)+3bc\left(b+c\right)+3ac\left(a+c\right)+6abc$$\Leftrightarrow ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\ge6abc$$\Leftrightarrow\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6$(Đúng theo Cô Si)"=" khi a=b=c=1

Nguyễn Trọng Chiến · Answer

281:Ta có:$ab+bc+ca=3abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3$$\dfrac{1}{\sqrt{a^3+b}}\le\dfrac{1}{\sqrt{2\sqrt{a^3b}}}=\dfrac{1}{\sqrt{2a}\cdot\sqrt[4]{ab}}\le\dfrac{1}{2\sqrt{2a}}\cdot\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\right)=\dfrac{1}{2\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\right)\le\dfrac{1}{2\sqrt{2}}\cdot\left[\dfrac{1}{a}+\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]=\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{a}+\dfrac{1}{2a}+\dfrac{1}{2b}\right)$ Chứng minh tương tự:$\dfrac{1}{\sqrt{b^3+c}}\le\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{b}+\dfrac{1}{2b}+\dfrac{1}{2c}\right);\dfrac{1}{\sqrt{c^3+a}}\le\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{c}+\dfrac{1}{2c}+\dfrac{1}{2a}\right)$$\Rightarrow\dfrac{1}{\sqrt{a^3+b}}+\dfrac{1}{\sqrt{b^3+c}}+\dfrac{1}{\sqrt{c^3+a}}\le\dfrac{1}{2\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{b}+\dfrac{1}{2b}+\dfrac{1}{2c}+\dfrac{1}{c}+\dfrac{1}{2c}+\dfrac{1}{2a}\right)=\dfrac{1}{2\sqrt{2}}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{3}{\sqrt{2}}$ Dấu = xảy ra $\Leftrightarrow a=b=c=1$Câu trả lời: 281:Ta có:$ab+bc+ca=3abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3$$\dfrac{1}{\sqrt{a^3+b}}\le\dfrac{1}{\sqrt{2\sqrt{a^3b}}}=\dfrac{1}{\sqrt{2a}\cdot\sqrt[4]{ab}}\le\dfrac{1}{2\sqrt{2a}}\cdot\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\right)=\dfrac{1}{2\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\right)\le\dfrac{1}{2\sqrt{2}}\cdot\left[\dfrac{1}{a}+\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]=\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{a}+\dfrac{1}{2a}+\dfrac{1}{2b}\right)$ Chứng minh tương tự:$\dfrac{1}{\sqrt{b^3+c}}\le\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{b}+\dfrac{1}{2b}+\dfrac{1}{2c}\right);\dfrac{1}{\sqrt{c^3+a}}\le\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{c}+\dfrac{1}{2c}+\dfrac{1}{2a}\right)$$\Rightarrow\dfrac{1}{\sqrt{a^3+b}}+\dfrac{1}{\sqrt{b^3+c}}+\dfrac{1}{\sqrt{c^3+a}}\le\dfrac{1}{2\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{b}+\dfrac{1}{2b}+\dfrac{1}{2c}+\dfrac{1}{c}+\dfrac{1}{2c}+\dfrac{1}{2a}\right)=\dfrac{1}{2\sqrt{2}}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{3}{\sqrt{2}}$ Dấu = xảy ra $\Leftrightarrow a=b=c=1$

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