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

Số hạng cuối là \(1+\dfrac{c}{2a}\) mới đúng chứ bạn?

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

\(E\ge3\sqrt[3]{\dfrac{a}{8b}}.3\sqrt[3]{\dfrac{b}{8c}}.3\sqrt[3]{\dfrac{c}{8a}}=\dfrac{27}{8}\)

\(E_{min}=\dfrac{27}{8}\) khi \(a=b=c\)

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

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 \(a=b=c\)

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

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

\(P=\dfrac{1}{a\left[2b+2c-\left(a+b+c\right)\right]}+\dfrac{1}{b\left[2c+2a-\left(a+b+c\right)\right]}+\dfrac{1}{c\left[2a+2b-\left(a+b+c\right)\right]}\)

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

\(P=\dfrac{1}{ab+ac-a^2}+\dfrac{1}{bc+ab-b^2}+\dfrac{1}{ca+bc-c^2}\)

Áp dụng bất đẳng thức cộng mẫu số

\(\Rightarrow P\ge\dfrac{\left(1+1+1\right)^2}{-a^2-b^2-c^2+2ab+2bc+2ca}=\dfrac{9}{-\left[a^2+b^2+c^2-2\left(ab+bc+ca\right)\right]}\) ( 1 )

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow a^2+b^2+c^2-2\left(ab+bc+ca\right)\ge-\left(ab+bc+ca\right)\)

\(\Rightarrow-\left[a^2+b^2+c^2-2\left(ab+bc+ca\right)\right]\le ab+bc+ca\)

\(\Rightarrow\dfrac{9}{-\left[a^2+b^2+c^2-2\left(ab+bc+ca\right)\right]}\ge\dfrac{9}{ab+bc+ca}\)

Từ ( 1 )

\(\Rightarrow P\ge\dfrac{9}{ab+bc+ca}\)

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow1\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{1}{3}\ge ab+bc+ca\)

\(\Rightarrow27\le\dfrac{9}{ab+bc+ca}\)

\(\Rightarrow P\ge27\)

Vậy \(P_{min}=27\)

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 dtsbn ta có:

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

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

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

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