ĐK:\(x,y,z\ge \frac{1}{2}\)

Cộng theo vế 3 BĐT trên ta có:

\(2x+2y+2z-\sqrt{4x-1}-\sqrt{4y-1}-\sqrt{4z-1}=0\)

\(\Leftrightarrow\left(4x-1-2\sqrt{4x-1}+1\right)+\left(4y-1-2\sqrt{4y-1}+1\right)+\left(4z-1-2\sqrt{4z-1}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{4x-1}-1\right)^2+\left(\sqrt{4y-1}-1\right)^2+\left(\sqrt{4z-1}-1\right)^2=0\)

Dễ thấy: \(VT\ge0\forall x,y,z\)

\("="\Leftrightarrow\left\{{}\begin{matrix}\sqrt{4x-1}=1\\\sqrt{4y-1}=1\\\sqrt{4z-1}=1\end{matrix}\right.\)\(\Leftrightarrow x=y=z=\dfrac{1}{2}\)

Lời giải:

ĐK \(x,y,z\geq \frac{1}{4}\)

\(\text{HPT}\Rightarrow 2(x+y+z)=\sqrt{4x-1}+\sqrt{4y-1}+\sqrt{4z-1}\)

Áp dụng bất đẳng thức AM-GM ta có :

\(\sqrt{4x-1}=\sqrt{(4x-1).1}\leq \frac{4x-1+1}{2}=2x\)

Tương tự với các biểu thức còn lại.....

\(\Rightarrow \sqrt{4x-1}+\sqrt{4y-1}+\sqrt{4z-1}\leq 2(x+y+z)\)

Dấu bằng xảy ra khi \(\left\{\begin{matrix} 4x-1=1\\ 4y-1=1\\ 4z-1=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{1}{2}\\ y=\frac{1}{2}\\ z=\frac{1}{2}\end{matrix}\right.\)

Vậy HPT có nghiệm \((x,y,z)=\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\)

x = y = z = 0,5

Đặt \(\left(x-1;y-2;z-3\right)=\left(a;b;c\right)=abc>0\)

Điều kiện bài toán trở thành :

\(a+1+b+2+c+3< 9\)

\(\sqrt{a+\sqrt{b}+\sqrt{c}}+\sqrt{c+5\left(a+1\right)+4\left(b+2\right)+3+\left(c+3\right)}\)

\(=\left(a+1\right)\left(b+2\right)=\left(b+2\right)\left(c+3\right)=\left(c+3\right)+\left(a+1\right)+11+a+b+c< 3\)

\(a+b+c< 3\)

\(=\sqrt{a+\sqrt{b}+\sqrt{c}+ab+bc+ca}\)

Mặt khác, do  không âm, ta luôn có:

\(\text{⇒2√a≥a(3−a)≥a(b+c)⇒2a≥a(3−a)≥a(b+c) (1)}\)

Hoàn toàn tương tự ta có:\(\text{ 2√b≥b(c+a)2b≥b(c+a) (2)}\)

\(\text{2√c≥c(a+b)2c≥c(a+b) (3)}\)

Cộng vế với vế (1);(2);(3):

Dấu "=" xảy ra khi và chỉ khi \(\text{a=b=c=0a=b=c=0 hoặc a=b=c=1a=b=c=1}\)

1. Với mọi số thực x;y;z ta có:

\(x^2+y^2+z^2+\dfrac{1}{2}\left(x^2+1\right)+\dfrac{1}{2}\left(y^2+1\right)+\dfrac{1}{2}\left(z^2+1\right)\ge xy+yz+zx+x+y+z\)

\(\Leftrightarrow\dfrac{3}{2}P+\dfrac{3}{2}\ge6\)

\(\Rightarrow P\ge3\)

\(P_{min}=3\) khi \(x=y=z=1\)

1.1

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=a>0\\\dfrac{1}{\sqrt{y}}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+\sqrt{2-b^2}=2\\b+\sqrt{2-a^2}=2\end{matrix}\right.\)

\(\Rightarrow a-b+\sqrt{2-b^2}-\sqrt{2-a^2}=0\)

\(\Leftrightarrow a-b+\dfrac{\left(a-b\right)\left(a+b\right)}{\sqrt{2-b^2}+\sqrt{2-a^2}}=0\)

\(\Leftrightarrow a=b\Leftrightarrow x=y\)

Thay vào pt đầu:

\(a+\sqrt{2-a^2}=2\Rightarrow\sqrt{2-a^2}=2-a\) (\(a\le2\))

\(\Leftrightarrow2-a^2=4-4a+a^2\Leftrightarrow2a^2-4a+2=0\)

\(\Rightarrow a=1\Rightarrow x=y=1\)

2.

\(\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^2-xy+y^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3xy+3y^2=21\\7x^2-7xy+7y^2=21\end{matrix}\right.\)

\(\Rightarrow4x^2-10xy+4y^2=0\)

\(\Leftrightarrow2\left(2x-y\right)\left(x-2y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\dfrac{1}{2}x\end{matrix}\right.\)

Thế vào pt đầu

...

DK : \(x,y,z\ge\frac{1}{2}\)

Cộng theo vế 3 BĐT trên ta có :

\(2x+2y+2z-\sqrt{4x-1}-\sqrt{4y-1}-\sqrt{4z-1}=0\)

\(\Leftrightarrow\left(4x-1-2\sqrt{4x-1}+1\right)+\left(4y-1-2\sqrt{4y-1}+1\right)\)

\(+\left(4z-1-2\sqrt{4z-1}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{4x-1}-1\right)^2+\left(\sqrt{4y-1}-1\right)^2+\left(\sqrt{4z-1}-1\right)^2=0\)

Dễ thấy : \(VT\ge0\forall x,y,z\)

" = " \(\Leftrightarrow\hept{\begin{cases}\sqrt{4x-1}=1\\\sqrt{4y-1}=1\\\sqrt{4z-1}=1\end{cases}\Leftrightarrow x=y=z=\frac{1}{2}}\)

Chúc bạn học tốt !!! 

ĐK: \(x,y,z\ge\frac{1}{4}\)

hệ pt <=> \(\hept{\begin{cases}x+y=\sqrt{4z-1}\\y+z=\sqrt{4x-1}\\z+x=\sqrt{4y-1}\end{cases}}\)

<=> \(\hept{\begin{cases}2x+2y=2\sqrt{4z-1}\\2y+2z=2\sqrt{4x-1}\\2z+2x=2\sqrt{4y-1}\end{cases}}\)

=> \(4x+4y+4z=2\sqrt{4z-1}+2\sqrt{4x-1}+2\sqrt{4y-1}\)

<=> \(\left(4x-1-2\sqrt{4x-1}+1\right)+\left(4y-1-2\sqrt{4y-1}+1\right)+\left(4z-1-2\sqrt{4z-1}+1\right)=0\)

<=> \(\left(\sqrt{4x-1}-1\right)^2+\left(\sqrt{4y-1}-1\right)^2+\left(\sqrt{4z-1}-1\right)^2=0\)

<=> \(\hept{\begin{cases}\sqrt{4x-1}-1=0\\\sqrt{4y-1}-1=0\\\sqrt{4z-1}-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}4x-1=1\\4y-1=1\\4z-1=1\end{cases}}\Leftrightarrow x=y=z=\frac{1}{2}\)(tm đk)

Thử vào thỏa mãn.

Vậy...

+ \(\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=4\Rightarrow x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=4\)

\(\Rightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)

+ \(x+1=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)\)

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

+ Tương tự : \(y+1=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\); \(z+1=\left(\sqrt{x}+\sqrt{z}\right)\left(\sqrt{y}+\sqrt{z}\right)\)

+ \(P=\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}+\sqrt{z}\right)^2\left(\sqrt{z}+\sqrt{x}\right)^2}\cdot\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{z}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)

\(=2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=2\)