Dấu >= hay <= vậy bạn? Bạn xem lại đề.

\(2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\right)\ge1+\dfrac{b}{b+1a}+\dfrac{c}{c+2b}+\dfrac{a}{a+2c}\)

\(\Leftrightarrow2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}+\dfrac{c}{a+2c}\right)\ge1+\dfrac{b+2a}{b+2a}+\dfrac{c+2b}{c+2b}+\dfrac{a+2c}{a+2c}=1+1+1+1=4\)Thật vậy:

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

\(\ge\dfrac{4a}{2\left(a+b+c\right)}+\dfrac{4b}{2\left(a+b+c\right)}+\dfrac{4c}{2\left(a+b+c\right)}=2\)

\(\Rightarrow VT\ge2.2=4\)

\(\RightarrowĐPCM\)

\(\dfrac{a^3}{\left(a+2b\right)\left(b+2c\right)}+\dfrac{a+2b}{27}+\dfrac{b+2c}{27}\ge3\sqrt[3]{\dfrac{a^3\left(a+2b\right)\left(b+2c\right)}{27^2.\left(a+2b\right)\left(b+2c\right)}}=\dfrac{a}{3}\)

Tương tự:

\(\dfrac{b^3}{\left(b+2c\right)\left(c+2a\right)}+\dfrac{b+2c}{27}+\dfrac{c+2a}{27}\ge\dfrac{b}{3}\)

\(\dfrac{c^3}{\left(c+2a\right)\left(a+2b\right)}+\dfrac{c+2a}{27}+\dfrac{a+2b}{27}\ge\dfrac{c}{3}\)

Cộng vế:

\(VT+\dfrac{2\left(a+b+c\right)}{9}\ge\dfrac{a+b+c}{3}\)

\(\Rightarrow VT\ge\dfrac{a+b+c}{9}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Leftrightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\text{ và }\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{a\cdot b\cdot c}{2a\cdot2b\cdot3c}=\dfrac{1}{8}\)

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)

\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)

=>\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)

b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)

=>\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)

Áp dụng t/c dtsbn ta có:

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

\(\dfrac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow2b=3a-c\)\(\dfrac{2c-b+a}{b}=2\Rightarrow2c-b+a=2b\Rightarrow2c=3b-a\)

\(\dfrac{2a+b-c}{c}=2\Rightarrow2a+b-c=2c\Rightarrow2a=3c-b\)

\(P=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{2a.2b.2c}=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{8abc}\)

b.\(ĐK:x;y\in Z^+;x;y\ne0\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{5}{x}+\dfrac{5}{y}=1\)

\(\Leftrightarrow\dfrac{5}{x}=1-\dfrac{5}{y}\)

\(\Leftrightarrow\dfrac{5}{x}=\dfrac{y-5}{y}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{y-5}\)

\(\Leftrightarrow x=\dfrac{5y}{y-5}\)

\(\Leftrightarrow x=5+\dfrac{25}{y-5}\) ( bạn chia \(5y\) cho \(y-5\) ý )

Để x;y là số nguyên dương thì \(25⋮y-5\) hay \(y-5\in U\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)

TH1: 

\(y-5=1\) 

\(\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=30\end{matrix}\right.\) ( tm )   ( bạn thế y=6 vào \(x=5+\dfrac{25}{y+5}\) nhé )

Xét tương tự, ta ra được nghiệm nguyên dương của phương trình:

\(\left\{{}\begin{matrix}x=30\\y=6\end{matrix}\right.\)  \(\left\{{}\begin{matrix}x=10\\y=10\end{matrix}\right.\)  \(\left\{{}\begin{matrix}x=6\\y=30\end{matrix}\right.\)

Áp dụng t/c dãy tỉ số bằng nhau ta có:

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

Do \(\dfrac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\)

\(\Rightarrow2b+c-a+a=3a\)

\(\Rightarrow2b+c=3a\Rightarrow3a-2b=c\)

Lại do \(\dfrac{2c-b+a}{b}=2\) \(\Rightarrow2c-b+a=2b\)

\(\Rightarrow2c+a-3b=0\)

\(\Rightarrow3b-2c=a\)

Ta lại có \(\dfrac{2a+b-c}{c}=2\Rightarrow2a+b-c=2c\)

\(\Rightarrow2a+b-c+c=3c\)

\(\Rightarrow2a +b=3c\)

\(\Rightarrow3c-2a=b\)

Khi đó:

\(P=\dfrac{c.a.b}{2b.2c.2a}=\dfrac{1}{8}\) (đoạn này mk làm hơi tắt, nếu không hiểu thì nói mk nhé!)

Vậy \(P=\dfrac{1}{8}.\)

Chú ý: Ở tử của p/s phải là 3a \(-2b\) mới làm được bài này.