Question

[Nhâm Gia Nghĩa]

Giả sử a , b , c là các số thực dương sao cho \(a+b+c=1\)

Chứng minh rằng \(2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\ge\dfrac{1+a}{1-a}+\dfrac{1+b}{1-b}+\dfrac{1+c}{1-c}\)

Answer 1

[Kuro Kazuya]

\(2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\ge\dfrac{1+a}{1-a}+\dfrac{1+b}{1-b}+\dfrac{1+c}{1-c}\)

Thay thế \(a+b+c=1\)

\(\Leftrightarrow2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\ge\dfrac{2a+b+c}{b+c}+\dfrac{a+2b+c}{a+c}+\dfrac{a+b+2c}{a+b}\)

\(\Leftrightarrow2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\ge\dfrac{2a}{b+c}+\dfrac{2b}{a+c}+\dfrac{2c}{a+b}+3\)

\(\Leftrightarrow\dfrac{2b}{a}+\dfrac{2c}{b}+\dfrac{2a}{c}\ge\dfrac{2a}{b+c}+\dfrac{2b}{a+c}+\dfrac{2c}{a+b}+3\)

\(\Leftrightarrow\left(\dfrac{2b}{a}-\dfrac{2b}{a+c}\right)+\left(\dfrac{2c}{b}-\dfrac{2c}{a+b}\right)+\left(\dfrac{2a}{c}-\dfrac{2a}{b+c}\right)\ge3\)

\(\Leftrightarrow\dfrac{2bc}{a\left(a+c\right)}+\dfrac{2ca}{b\left(a+b\right)}+\dfrac{2ab}{c\left(b+c\right)}\ge3\)

\(\Leftrightarrow\dfrac{bc}{a\left(a+c\right)}+\dfrac{ca}{b\left(a+b\right)}+\dfrac{ab}{c\left(b+c\right)}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\left(bc\right)^2}{abc\left(a+c\right)}+\dfrac{\left(ca\right)^2}{abc\left(a+b\right)}+\dfrac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\dfrac{3}{2}\)

Áp dụng bất đẳng thức cộng mẫu số

\(\Rightarrow\dfrac{\left(bc\right)^2}{abc\left(a+c\right)}+\dfrac{\left(ca\right)^2}{abc\left(a+b\right)}+\dfrac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\dfrac{\left(ab+bc+ca\right)^2}{abc\left(a+b+c+a+b+c\right)}=\dfrac{\left(ab+bc+ca\right)^2}{2abc}\)

Chứng minh rằng \(\dfrac{\left(ab+bc+ca\right)^2}{2abc}\ge\dfrac{3}{2}\)

\(\Leftrightarrow2\left(ab+bc+ca\right)^2\ge6abc\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc\ge3abc\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\ge3abc\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\ge3abc\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\)

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow\left\{{}\begin{matrix}a^2b^2+b^2c^2\ge2\sqrt{a^2b^4c^2}=2ab^2c\\b^2c^2+c^2a^2\ge2\sqrt{a^2b^2c^4}=2abc^2\\a^2b^2+c^2a^2\ge2\sqrt{a^4b^2c^2}=2a^2bc\end{matrix}\right.\)

\(\Leftrightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)\ge2abc\left(a+b+c\right)\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\) ( đpcm )

Vì \(\dfrac{\left(ab+bc+ca\right)^2}{2abc}\ge\dfrac{3}{2}\)

Vậy \(\dfrac{\left(bc\right)^2}{abc\left(a+c\right)}+\dfrac{\left(ca\right)^2}{abc\left(a+b\right)}+\dfrac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\dfrac{3}{2}\)

\(\Leftrightarrow2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\ge\dfrac{1+a}{1-a}+\dfrac{1+b}{1-b}+\dfrac{1+c}{1-c}\)( đpcm )