Chứng minh 1/(2a+b)^2+1/(2b+c)^2+1/(2c+a)^2≥1/ab+bc+ca

Bài 2: Ta có

\(\frac{1}{\left(2a+b\right)^2}+\frac{1}{\left(2b+c\right)^2}+\frac{1}{\left(2c+a\right)^2}=\frac{\left(2c+b\right)^2}{\left(2a+b\right)^2\left(2c+b\right)^2}+\frac{\left(2a+c\right)^2}{\left(2b+c\right)^2\left(2a+c\right)^2}+\frac{\left(2b+a\right)^2}{\left(2c+a\right)^2\left(2b+a\right)^2}\ge\frac{9\left(a+b+c\right)^2}{\left(2a+b\right)^2\left(2c+b\right)^2+\left(2b+c\right)^2\left(2a+c\right)^2+\left(2c+a\right)^2\left(2b+a\right)^2}\)

Ta đặt: \(\left\{\begin{matrix}a+b+c=m\\ab+bc+ca=n\end{matrix}\right.\)

Ta có: \(\left\{\begin{matrix}\left(2a+b\right)\left(2c+b\right)=b^2+2ac+2\left(ab+bc+ca\right)=b^2+2ca+2n\\\left(2b+c\right)\left(2a+c\right)=c^2+2ab+2n\\\left(2c+a\right)\left(2b+a\right)=a^2+2bc+2n\end{matrix}\right.\)

Từ đây ta suy ra: \(\left(2a+b\right)^2\left(2c+b\right)^2+\left(2b+c\right)^2\left(2a+c\right)^2+\left(2c+a\right)^2\left(2b+a\right)^2=\left(b^2+2ca+2n\right)^2+\left(c^2+2ab+2n\right)^2+\left(a^2+2bc+2n\right)^2\)

\(=12n^2+4n\left(a^2+b^2+c^2+2ab+2bc+2ca\right)+\left(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\right)+2\left(a^2b^2+b^2c^2+c^2a+2a^2bc+2ab^2c+2abc^2\right)\)

\(=12n^2+4n\left(a+b+c\right)^2+\left(a^2+b^2+c^2\right)^2+2\left(ab+bc+ca\right)^2\)

\(=14n^2+4nm^2+\left(m^2-2n\right)^2\)

\(=18n^2+m^4\)

Vậy ta cần chứng minh: \(\frac{9m^2}{18n+m^4}\ge\frac{1}{n}\)

\(\Leftrightarrow9m^2n-18n^2-m^4\ge0\)

\(\Leftrightarrow\left(m^2-3n\right)\left(6n-m^2\right)\ge0\)

Đễ thấy \(m^2\ge3n\) vì \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

Giờ ta chỉ cần xét 2 trường hợp:

Trường hợp 1: \(6n\ge m^2\) với trường hợp này thì ta có ĐPCM

Trường hợp 2: \(6n< m^2\). Ta giả sử a là số lớn nhất trong 3 số a, b, c

\(\Leftrightarrow6\left(ab+bc+ca\right)< \left(a+b+c\right)^2\)

\(\Leftrightarrow a^2+b^2+c^2-4\left(ab+bc+ca\right)>0\)

\(\Leftrightarrow a^2-4a\left(b+c\right)+4\left(b+c\right)^2-3b^2-3c^2-12bc>0\)

\(\Leftrightarrow\left(a-2b-2c\right)^2>3\left(b^2+c^2+4bc\right)\)

\(\Leftrightarrow\left[\begin{matrix}a-2b-2c>\sqrt{3\left(b^2+c^2+4bc\right)}\\a-2b-2c< -\sqrt{3\left(b^2+c^2+4bc\right)}\end{matrix}\right.\)

Xét trường hợp: \(a-2b-2c< -\sqrt{3\left(b^2+c^2+4bc\right)}\)

\(\Leftrightarrow a< 2b+2c-\sqrt{3\left(b^2+c^2+4bc\right)}< \left(2-\sqrt{3}\right)\left(b+c\right)< \frac{b+c}{2}\left(l\right)\)

Xét trường hợp: \(a-2b-2c>\sqrt{3\left(b^2+c^2+4bc\right)}\)

\(\Leftrightarrow a>2\left(b+c\right)+\sqrt{3\left(b^2+c^2+4bc\right)}>\left(2+\sqrt{3}\right)\left(b+c\right)\)

Giờ ta quay lại bài toán ban đầu:

Ta có: \(\frac{1}{\left(2a+b\right)^2}+\frac{1}{\left(2c+a\right)^2}\ge\frac{2}{\left(2a+b\right)\left(2c+a\right)}>\frac{2}{2\left(a+2b\right)\left(a+2c\right)}\ge\frac{4}{\left(a+2b+a+2c\right)^2}=\frac{1}{\left(a+b+c\right)^2}\)

Và \(\frac{1}{\left(2b+c\right)^2}>\frac{1}{4\left(b+c\right)^2}\)

Từ đây ta có:

\(\frac{1}{\left(2a+b\right)^2}+\frac{1}{\left(2b+c\right)^2}+\frac{1}{\left(2c+a\right)^2}>\frac{1}{\left(a+b+c\right)^2}+\frac{1}{4\left(b+c\right)^2}\)

Giờ ta phải chứng minh: \(\frac{1}{\left(a+b+c\right)^2}+\frac{1}{4\left(b+c\right)^2}>\frac{1}{ab+ac}>\frac{1}{ab+bc+ca}\)

\(\Leftrightarrow\frac{1}{\left(\frac{a}{b+c}+1\right)^2}+\frac{1}{4}>\frac{b+c}{a}\)

Ta đặt \(t=\frac{a}{b+c}>2+\sqrt{3}\) thì ta có

\(\Leftrightarrow\frac{1}{\left(t+1\right)^2}+\frac{1}{4}-\frac{1}{t}>0\)

\(\Leftrightarrow t^3-2t^2-3t-4>0\)

\(\Leftrightarrow\left(t-2-\sqrt{3}\right)\left(t^2+\sqrt{3}t+2\sqrt{3}\right)+4\sqrt{3}+2>0\) (đúng)

Vậy ta có ĐPCM