Áp dụng BĐt cô-si, ta có \(\frac{2\left(a+b\right)^2}{2a+3b}\ge\frac{8ab}{2a+3b}=\frac{8}{\frac{2}{b}+\frac{3}{a}}\)

                                      \(\frac{\left(b+2c\right)^2}{2b+c}\ge\frac{8bc}{2b+c}=\frac{8}{\frac{2}{c}+\frac{1}{b}}\)

                                        \(\frac{\left(2c+a\right)^2}{c+2a}\ge\frac{8ac}{c+2a}\ge\frac{8}{\frac{1}{a}+\frac{2}{c}}\)

Cộng 3 cái vào, ta có 

A\(\ge8\left(\frac{1}{\frac{2}{b}+\frac{3}{a}}+\frac{1}{\frac{1}{b}+\frac{2}{c}}+\frac{1}{\frac{1}{a}+\frac{2}{c}}\right)\ge8\left(\frac{9}{\frac{3}{b}+\frac{4}{c}+\frac{4}{a}}\right)=8.\frac{9}{3}=24\)

Vậy A min = 24 

Neetkun ^^

bạn tìm ra dấu= xảy ra khi nào

rất tiếc sai rồi :)) 

Vì \(a,b,c>0\Rightarrow a+b+c\ne0\)

Áp dụng tc dtsbn:

\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Rightarrow P=\dfrac{abc}{2a\cdot2b\cdot2c}=\dfrac{1}{8}\)

\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)

Tương tự và cộng lại:

\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)

\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)

\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)

Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z

\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)

Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))

\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))

\(\Rightarrow P\le\dfrac{3}{16}\)

\(ĐTXR\Leftrightarrow a=b=c=1\)

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:

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

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

Trước hết theo BĐT Schur bậc 3 ta có:

\(\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc\ge2\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+3abc\ge2\left(ab+bc+ca\right)\) (do \(a+b+c=3\)) (1)

Đặt vế trái BĐT cần chứng minh là P, ta có:

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

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

Áp dụng (1):

\(\Rightarrow P\ge\dfrac{\left[2\left(ab+bc+ca\right)\right]^2}{\left(ab+bc+ca\right)^2}=4\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

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