\(ĐK:x,y,z>\frac{1}{2}\)

Ta có: \(\left(x+2y\right)^2=\left(\frac{3y}{2}+\frac{y+2x}{2}\right)^2\ge4.\frac{3y}{2}.\frac{y+2x}{2}=3y\left(2x+y\right)\)\(\Rightarrow\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\Rightarrow\frac{2x+y}{x\left(x+2y\right)}\le\frac{x+2y}{3xy}=\frac{1}{3}\left(\frac{2}{x}+\frac{1}{y}\right)\)

Tương tự: \(\frac{2y+z}{y\left(y+2z\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\); \(\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\)

Cộng theo vế ba bất đẳng thức trên, ta được: \(VT\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)

Đẳng thức xảy ra khi x = y = z = 1

a, \(\hept{\begin{cases}x-y=5\\xy=6\end{cases}}\Rightarrow\hept{\begin{cases}x=5+y\left(1\right)\\xy=6\left(2\right)\end{cases}}\)

Thay (1) vào (2) ta được : 

\(\left(5+y\right)y=6\Leftrightarrow5y+y^2-6=0\)

\(\Leftrightarrow\left(y-1\right)\left(y+6\right)=0\Leftrightarrow y=1;-6\)

TH1 : thay y = 1 vào (1) ta được : \(x=5+1=6\)

TH2 : thay y = -6 vào (1) ta được : \(x=5+\left(-6\right)=5-6=-1\)

đk: \(x\ge3\)

Ta có: \(\sqrt{x^2-3x}-\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x\left(x-3\right)}-\sqrt{x-3}=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\sqrt{x-3}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1=0\\\sqrt{x-3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=1\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=3\left(tm\right)\end{cases}}\)

Vậy x = 3

\(\sqrt{x^2-3x}-\sqrt{x-3}=0\)ĐKXĐ : \(x\ge3\)

\(\Leftrightarrow\sqrt{x\left(x-3\right)}-\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x}\sqrt{x-3}-\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)

TH1 : \(\sqrt{x-3}=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)

TH2 : \(\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)ktm do \(x\ge3\)

Đặt số cần tìm là \(\overline{ab},\left(0\le a,b\le9;a,b\inℕ;a\ne0,a+b=8\right)\)

Số sau khi đổi vị trí là \(\overline{ba}\).

Theo bài ra ta có: \(\overline{ab}-\overline{ba}=18\Leftrightarrow10a+b-\left(10b+a\right)=18\Leftrightarrow9a-9b=18\Leftrightarrow a-b=2\)

\(\Rightarrow a-\left(8-a\right)=2\Leftrightarrow2a=10\Leftrightarrow a=5\Rightarrow b=3\)(thỏa) 

Áp dụng bđt Bunyakovsky dạng phân thức ta có :

\(P=\left(\frac{a}{a+b}\right)^4+\left(\frac{b}{b+c}\right)^4+\left(\frac{c}{c+a}\right)^4\ge\frac{\left[\left(\frac{a}{a+b}\right)^2+\left(\frac{b}{b+c}\right)^2+\left(\frac{c}{c+a}\right)^2\right]^2}{3}\)(1)

Tiếp tục sử dụng bđt Bunyakovsky dạng phân thức ta có :

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

Đặt  \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)

Áp dụng bđt Cauchy ta có :

\(\frac{a}{a+b}+\frac{a+b}{4a}\ge2\sqrt{\frac{a}{a+b}\cdot\frac{a+b}{4a}}=1\)

=> \(A+\frac{a+b}{4a}+\frac{b+c}{4b}+\frac{c+a}{4c}\ge3\)

=> \(A+\frac{a}{4a}+\frac{b}{4a}+\frac{b}{4b}+\frac{c}{4b}+\frac{c}{4c}+\frac{a}{4c}\ge3\)

=> \(A+\frac{3}{4}+\frac{b}{4a}+\frac{c}{4b}+\frac{a}{4c}\ge3\)

Theo Cauchy ta có : \(\frac{b}{4a}+\frac{c}{4b}+\frac{a}{4c}\ge3\sqrt[3]{\frac{b}{4a}\cdot\frac{c}{4b}+\frac{a}{4c}}=\frac{3}{4}\)

=> \(A+\frac{3}{4}+\frac{3}{4}\ge3\)=> \(A\ge\frac{3}{2}\)(3)

Từ (1), (2) và (3) => \(P\ge\frac{3}{16}\)

Đẳng thức xảy ra <=> a = b = c

Vậy MinP = 3/16 <=> a = b = c 

Ta có:

\(P=\left(\frac{a}{a+b}\right)^4+\left(\frac{b}{b+c}\right)^4+\left(\frac{c}{c+a}\right)^4=\left(\frac{1}{1+\frac{b}{a}}\right)^4+\left(\frac{1}{1+\frac{c}{b}}\right)^4+\left(\frac{1}{1+\frac{a}{c}}\right)^4\)

Đặt \(\left(\frac{b}{a},\frac{c}{b},\frac{a}{c}\right)=\left(x,y,z\right)\left(x,y,z>0\right)\) \(\Rightarrow xyz=1\)

Khi đó: \(P=\frac{1}{\left(1+x\right)^4}+\frac{1}{\left(1+y\right)^4}+\frac{1}{\left(1+z\right)^4}\)

\(\ge3\left[\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}+\frac{1}{\left(1+z\right)^2}\right]^2\)

Ta có: \(\left(1+xy\right)\left(1+\frac{x}{y}\right)\ge\left(1+x\right)^2\Leftrightarrow\left(1+x\right)^2\le\frac{\left(1+xy\right)\left(x+y\right)}{y}\)( Bunyakovsky)

\(\Leftrightarrow\frac{1}{\left(1+x\right)^2}\ge\frac{y}{\left(1+xy\right)\left(x+y\right)}\) ; tương tự: \(\frac{1}{\left(1+y\right)^2}\ge\frac{x}{\left(1+xy\right)\left(x+y\right)}\)

Áp dụng BĐT Cauchy: \(\frac{1}{\left(1+z\right)^2}+\frac{1}{4}\ge2\sqrt{\frac{1}{\left(1+z\right)^2}\cdot\frac{1}{4}}=\frac{1}{1+z}\)

\(\Rightarrow\frac{1}{\left(1+z\right)^2}\ge\frac{1}{1+z}-\frac{1}{4}\)

Khi đó: \(P\ge\frac{1}{3}\left[\frac{x}{\left(1+xy\right)\left(x+y\right)}+\frac{y}{\left(1+xy\right)\left(x+y\right)}+\frac{1}{1+z}-\frac{1}{4}\right]^2\)

\(=\frac{1}{3}\left(\frac{1}{1+xy}+\frac{1}{1+z}-\frac{1}{4}\right)^2=\frac{1}{3}\left(\frac{xyz}{xyz+xy}+\frac{1}{1+z}-\frac{1}{4}\right)^2\)

\(=\frac{1}{3}\left(\frac{z}{1+z}+\frac{1}{1+z}-\frac{1}{4}\right)^2=\frac{1}{3}\left(1-\frac{1}{4}\right)^2=\frac{3}{16}\)

Dấu "=" xảy ra khi: a = b = c

Vậy Min(P) = 3/16 khi a = b = c