\(\dfrac{a+b}{7}=\dfrac{a-b}{-2}=\dfrac{2a}{5}\\ \Rightarrow-4a=5a-5b\Rightarrow9a=5b\Rightarrow\dfrac{a}{5}=\dfrac{b}{9}\\ \text{Đặt }\dfrac{a}{5}=\dfrac{b}{9}=k\Rightarrow a=5k;b=9k\\ \Rightarrow a^2=25k^2;b^2=81k^2\\ \Rightarrow a^2< b^2\left(25< 81\right)\)

\(\dfrac{S_{GFH}}{S_{AGE}}=\dfrac{\dfrac{1}{2}GF\cdot GH}{\dfrac{1}{2}AG\cdot GE}=\dfrac{\dfrac{1}{2}\cdot\dfrac{1}{2}AG\cdot\dfrac{1}{2}GE}{\dfrac{1}{2}AG\cdot GE}=\dfrac{1}{4}\\ \Rightarrow S_{GFH}=\dfrac{1}{4}S_{AGE}\left(1\right)\\ \dfrac{S_{AGE}}{S_{ADE}}=\dfrac{GE}{DE}=\dfrac{GH+HE}{DH+HG+HE}=\dfrac{2DH}{3DH}=\dfrac{2}{3}\\ \Rightarrow S_{AGE}=\dfrac{2}{3}S_{ADE}\left(2\right)\\ \dfrac{S_{ADE}}{S_{ADC}}=\dfrac{AE}{AC}=\dfrac{4}{5}\Rightarrow S_{ADE}=\dfrac{4}{5}S_{ADC}\left(3\right)\\ \dfrac{S_{ADC}}{S_{ABC}}=\dfrac{DC}{BC}=\dfrac{5}{6}\Rightarrow S_{ADC}=\dfrac{5}{6}S_{ABC}\left(4\right)\)

\(\left(1\right)\left(2\right)\left(3\right)\left(4\right)\Rightarrow S_{GFH}=\dfrac{1}{4}\cdot\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot S_{ABC}=\dfrac{1}{9}S_{ABC}\)

Cộng VTV: 

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\ge\dfrac{4}{2a+b+c}+\dfrac{4}{a+2b+c}+\dfrac{4}{a+b+2c}\\ \Leftrightarrow2\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge4\left(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\right)\\ \Leftrightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge2\left(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\right)\)

best mục tiêu: tìm đc ông thầy NHM :)

\(VT=\left(\dfrac{b}{a}+\dfrac{b}{c}\right)+\left(\dfrac{c}{a}+\dfrac{c}{b}\right)+\left(\dfrac{a}{b}+\dfrac{a}{c}\right)\)

Ta có \(\left(\dfrac{b}{c}+\dfrac{b}{a}\right)\left(a+c\right)\ge\left(\sqrt{b}+\sqrt{b}\right)^2=4b\Leftrightarrow\dfrac{b}{c}+\dfrac{b}{a}\ge\dfrac{4b}{a+c}\)

CMTT \(\Leftrightarrow\left(\dfrac{c}{a}+\dfrac{c}{b}\right)\ge\dfrac{4c}{a+b};\dfrac{a}{b}+\dfrac{a}{c}\ge\dfrac{4a}{b+c}\)

Cộng VTV ta đc đpcm

Dấu \("="\Leftrightarrow a=b=c\)

Ta có \(\dfrac{1}{a+b}+\dfrac{1}{b+c}\ge\dfrac{4}{a+2b+c}\)

\(\Rightarrow VT\ge2\left(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\right)\left(1\right)\)

Ta cần cm \(\dfrac{1}{2a+b+c}\ge\dfrac{4}{a^2+28}\)

\(\Leftrightarrow a^2+28\ge4a+8b+4c\\ \Leftrightarrow2a^2+b^2+c^2+16-4b-8a-4c\ge0\\ \Leftrightarrow2\left(a-2\right)^2+\left(b-2\right)^2+\left(c-2\right)^2\ge0\left(\text{đúng}\right)\)

\(\Leftrightarrow\sum\dfrac{1}{2a+b+c}\ge\sum\dfrac{8}{a^2+28}\)

Kết hợp \(\left(1\right)\) ta đc đpcm

Dấu \("="\Leftrightarrow a=b=c=2\)

Bài 6:

Đặt \(11...11\left(\text{n chữ số }1\right)=a\left(a\in N\right)\)

\(\Rightarrow22...22=2a;44...44=4a;66...66=6a\\ 10...00\left(n\text{ chữ số }0\right)=9a+1\)

\(\Rightarrow S=44....44\cdot10...00\left(\text{n chữ số }0\right)+22....22-66...66\\ \Rightarrow S=4a\cdot\left(9a+1\right)+2a-6a=36a^2+4a+2a-6a=36a^2=\left(6a\right)^2\left(đpcm\right)\)