Câu hỏi của Phạm Vũ Lam Khánh

Question

Giả sử a , b , c là các số thực dương sao cho a+b+c=1Chứng minh rằng : $2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}$

Kudo Shinichi · Accepted Answer

$2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}$Thay thế $a+b+c=1$$\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{2a+b+c}{b+c}+\frac{a+2b+c}{a+c}+\frac{a+b+2c}{a+b}$$\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}+3$$\Leftrightarrow\frac{2b}{a}+\frac{2c}{b}+\frac{2a}{c}\ge\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}+3$$\Leftrightarrow\left(\frac{2b}{a}-\frac{2b}{a+c}\right)+\left(\frac{2c}{b}-\frac{2c}{a+b}\right)+\left(\frac{2a}{c}-\frac{2a}{b+c}\right)\ge3$$\Leftrightarrow\frac{2bc}{a\left(a+c\right)}+\frac{2ca}{b\left(a+b\right)}+\frac{2ab}{c\left(b+c\right)}\ge3$$\Leftrightarrow\frac{bc}{a\left(a+c\right)}+\frac{ca}{b\left(a+b\right)}+\frac{ab}{c\left(b+c\right)}\ge\frac{3}{2}$$\Leftrightarrow\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\frac{3}{2}$Áp dụng bất đẳng thức cộng mẫu số $\Rightarrow\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}$$\ge\frac{\left(ab+bc+ca\right)^2}{abc\left(a+b+c+a+b+c\right)}=\frac{\left(ab+bc+ca\right)^2}{2abc}$Chứng minh rằng : $\frac{\left(ab+bc+ca\right)^2}{2abc}\ge\frac{3}{2}$$\Leftrightarrow2\left(ab+bc+ca\right)^2\ge6abc$$\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc\ge3abc$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\ge3abc$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\ge3abc$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc$Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm$\Rightarrow\hept{\begin{cases}a^2b^2+b^2c^2\ge2\sqrt{a^2b^4c^2}=2ab^2c\b^2c^2+c^2a^2\ge2\sqrt{a^2b^2c^4}=2abc^2\a^2b^2+c^2a^2\ge2\sqrt{a^2b^2c^2}=2a^2bc\end{cases}}$$\Leftrightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)\ge2abc\left(a+b+c\right)$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\left(đpcm\right)$Vì $\frac{\left(ab+bc+ca\right)^2}{2abc}\ge\frac{3}{2}$Vậy $\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\frac{3}{2}$$\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\left(đpcm\right)$Chúc bạn học tốt !!!Câu trả lời: $2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}$Thay thế $a+b+c=1$$\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{2a+b+c}{b+c}+\frac{a+2b+c}{a+c}+\frac{a+b+2c}{a+b}$$\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}+3$$\Leftrightarrow\frac{2b}{a}+\frac{2c}{b}+\frac{2a}{c}\ge\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}+3$$\Leftrightarrow\left(\frac{2b}{a}-\frac{2b}{a+c}\right)+\left(\frac{2c}{b}-\frac{2c}{a+b}\right)+\left(\frac{2a}{c}-\frac{2a}{b+c}\right)\ge3$$\Leftrightarrow\frac{2bc}{a\left(a+c\right)}+\frac{2ca}{b\left(a+b\right)}+\frac{2ab}{c\left(b+c\right)}\ge3$$\Leftrightarrow\frac{bc}{a\left(a+c\right)}+\frac{ca}{b\left(a+b\right)}+\frac{ab}{c\left(b+c\right)}\ge\frac{3}{2}$$\Leftrightarrow\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\frac{3}{2}$Áp dụng bất đẳng thức cộng mẫu số $\Rightarrow\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}$$\ge\frac{\left(ab+bc+ca\right)^2}{abc\left(a+b+c+a+b+c\right)}=\frac{\left(ab+bc+ca\right)^2}{2abc}$Chứng minh rằng : $\frac{\left(ab+bc+ca\right)^2}{2abc}\ge\frac{3}{2}$$\Leftrightarrow2\left(ab+bc+ca\right)^2\ge6abc$$\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc\ge3abc$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\ge3abc$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\ge3abc$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc$Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm$\Rightarrow\hept{\begin{cases}a^2b^2+b^2c^2\ge2\sqrt{a^2b^4c^2}=2ab^2c\b^2c^2+c^2a^2\ge2\sqrt{a^2b^2c^4}=2abc^2\a^2b^2+c^2a^2\ge2\sqrt{a^2b^2c^2}=2a^2bc\end{cases}}$$\Leftrightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)\ge2abc\left(a+b+c\right)$$\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\left(đpcm\right)$Vì $\frac{\left(ab+bc+ca\right)^2}{2abc}\ge\frac{3}{2}$Vậy $\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\frac{3}{2}$$\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\left(đpcm\right)$Chúc bạn học tốt !!!

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