Ta có: \(x+y+z=1\Rightarrow\hept{\begin{cases}\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\\\sqrt{y+xz}=\sqrt{y\left(x+y+z\right)+xz}=\sqrt{\left(x+y\right)\left(y+z\right)}\\\sqrt{z+xy}=\sqrt{z\left(x+y+z\right)+xy}=\sqrt{\left(x+z\right)\left(y+z\right)}\end{cases}}\)

Ta viết lại A

\(A=\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(y+z\right)\left(x+z\right)}\)

Áp dụng bđt AM-GM:

\(A\le\frac{x+y+x+z+x+y+y+z+y+z+x+z}{2}=2\)

\("="\Leftrightarrow x=y=z=\frac{1}{3}\)

\(x+yz=x\left(x+y+z\right)+yz\)

\(=x^2+xy+xz+yz\)

\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)

+ Tương tự : \(y+xz=\left(x+y\right)\left(y+z\right)\)

\(z+xy=\left(x+z\right)\left(y+z\right)\)

+ Theo bđt AM-GM : \(\sqrt{\left(x+y\right)\left(x+z\right)}\le\frac{x+y+x+z}{2}\)

\(\Rightarrow\sqrt{\left(x-1\right)\left(y-1\right)}\le\frac{2x+y+z}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x+y=x+z\Leftrightarrow y=z\)

+ Tương tự ta cm đc : 

\(\sqrt{\left(x+y\right)\left(y+z\right)}\le\frac{x+2y+z}{2}\).   Dấu "=" xảy ra \(\Leftrightarrow x=z\)

\(\sqrt{\left(x+z\right)\left(y+z\right)}\le\frac{x+y+2z}{2}\).   Dấu "=" xảy ra \(\Leftrightarrow x=y\)

Do đó : \(A\le\frac{4\left(x+y+z\right)}{2}=2\)

A = 2 \(\Leftrightarrow x=y=z=\frac{1}{3}\)

Vậy Max A = 2 \(\Leftrightarrow x=y=z=\frac{1}{3}\)

x+yz=x(x+y+z)+yz

=x2+xy+xz+yz

=x(x+y)+z(x+y)=(x+z)(x+y)

+ Tương tự : y+xz=(x+y)(y+z)

z+xy=(x+z)(y+z)

+ Theo bđt AM-GM : √(x+y)(x+z)≤x+y+x+z2 

⇒√(x−1)(y−1)≤2x+y+z2 

Dấu "=" xảy ra ⇔x+y=x+z⇔y=z

+ Tương tự ta cm đc : 

√(x+y)(y+z)≤x+2y+z2 .   Dấu "=" xảy ra ⇔x=z

√(x+z)(y+z)≤x+y+2z2 .   Dấu "=" xảy ra ⇔x=y

Do đó : A≤4(x+y+z)2 =2

A = 2 ⇔x=y=z=13 

Vậy Max A = 2 

\(\sqrt{4x+2\sqrt{x}+1}\le\sqrt{4x+\dfrac{1}{2}\left(2^2+x\right)+1}=\sqrt{\dfrac{9x}{2}+3}\)

\(=\dfrac{1}{\sqrt{21}}.\sqrt{21}.\sqrt{\dfrac{9x}{2}+3}\le\dfrac{1}{2\sqrt{21}}\left(21+\dfrac{9x}{2}+3\right)=\dfrac{1}{2\sqrt{21}}\left(\dfrac{9x}{2}+24\right)\)

Tương tự và cộng lại:

\(A\le\dfrac{1}{2\sqrt{21}}\left(\dfrac{9}{2}\left(x+y+z\right)+72\right)=3\sqrt{21}\)

\(A_{max}=3\sqrt{21}\) khi \(x=y=z=4\)

\(A=1\sqrt{4x+2\sqrt{x}+1}+1.\sqrt{4y+2\sqrt{y}+1}+1\sqrt{4z+2\sqrt{z}+1}\)

\(\le\sqrt{\left(1+1+1\right)\left(4\left(x+y+z\right)+2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3\right)}\)

\(=\sqrt{3.\left[51+\dfrac{4\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}{2}\right]}\)

\(\le\sqrt{3.\left[51+\dfrac{x+y+z+12}{2}\right]}\)

\(=\sqrt{189}\)

Dấu "=" xảy ra <=> x = y = z = 4

không biết :))))

áp dụng bunhiacopski ta có: 

P^2 =< (1+1+1)(1/1+x^2 + 1/1+y^2+1/1+z^2)= 3(....)

đặt (...) =A

ta có: 1/1+x^2=< 1/2x

tt với 2 cái kia

=> A=< 1/2(1/x+1/y+1/z) =<1/2 ( xy+yz+xz / xyz)=1/2 ..........

đoạn sau chj chịu

^^ sorry

Bài này là câu lớp 8 rất quen thuộc rùiiiiiii !!!!!!!!

gt <=>    \(\frac{x+y+z}{xyz}=1\)

<=>    \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)

Đặt:   \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)

=>    \(ab+bc+ca=1\)

VÀ:    \(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\)

THAY VÀO P TA ĐƯỢC:    

\(P=\frac{1}{\sqrt{1+\frac{1}{a^2}}}+\frac{1}{\sqrt{1+\frac{1}{b^2}}}+\frac{1}{\sqrt{1+\frac{1}{c^2}}}\)

=>     \(P=\frac{1}{\sqrt{\frac{a^2+1}{a^2}}}+\frac{1}{\sqrt{\frac{b^2+1}{b^2}}}+\frac{1}{\sqrt{\frac{c^2+1}{c^2}}}\)

=>     \(P=\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\)

Thay     \(1=ab+bc+ca\)    vào P ta sẽ được:

=>      \(P=\frac{a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)

=>     \(P=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(b+a\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)

=>      \(2P=2.\sqrt{\frac{a}{a+b}}.\sqrt{\frac{a}{a+c}}+2.\sqrt{\frac{b}{b+a}}.\sqrt{\frac{b}{b+c}}+2.\sqrt{\frac{c}{c+a}}.\sqrt{\frac{c}{c+b}}\)

TA ÁP DỤNG BĐT CAUCHY 2 SỐ SẼ ĐƯỢC:

=>      \(2P\le\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{b+a}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{c}{c+b}\)

=>     \(2P\le\left(\frac{a}{a+b}+\frac{b}{b+a}\right)+\left(\frac{b}{b+c}+\frac{c}{c+b}\right)+\left(\frac{c}{c+a}+\frac{a}{a+c}\right)\)

=>     \(2P\le\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\)

=>     \(2P\le1+1+1=3\)

=>     \(P\le\frac{3}{2}\)

DẤU "=" XẢY RA <=>    \(a=b=c\)    . MÀ     \(ab+bc+ca=1\)

=>     \(a=b=c=\sqrt{\frac{1}{3}}\)

=>     \(x=y=z=\sqrt{3}\)

VẬY P MAX \(=\frac{3}{2}\)      <=>      \(x=y=z=\sqrt{3}\)

Lời giải:

Áp dụng BĐT AM-GM:

$y\sqrt{x-1}=\sqrt{y^2(x-1)}=\sqrt{y(xy-y)}\leq \frac{y+xy-y}{2}=\frac{xy}{2}$

$x\sqrt{y-2}=\sqrt{x^2(y-2)}=\sqrt{x(xy-2x)}\leq \frac{2x+(xy-2x)}{2\sqrt{2}}=\frac{xy}{2\sqrt{2}}$

$\Rightarrow y\sqrt{x-1}+x\sqrt{y-2}\leq \frac{xy}{2}+\frac{xy}{2\sqrt{2}}=xy.\frac{2+\sqrt{2}}{4}$

$\Rightarrow P\leq \frac{2+\sqrt{2}}{4}$

Vậy $P_{\max}=\frac{2+\sqrt{2}}{4}$

\(A=\sum\sqrt{4x+2\sqrt{x}+1}\)

\(Max_A=+\infty\)

\("="x=y=z=+\infty\)

Các biểu thức ở trong căn đều đưa được về bình phương
\(\sqrt{4x+2\sqrt{x}+1}=\sqrt{\left(2\sqrt{x}+1\right)^2}=\left|2\sqrt{x}+1\right|=2\sqrt{x}+1\)

Tương tự với hai căn còn lại ta sẽ có biểu thức đề cho tương đương với
\(2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3\)