ĐKXĐ : \(\left\{{}\begin{matrix}x>2019\\y>2020\\z>2021\end{matrix}\right.\)

Đặt \(\sqrt{x-2019}=a,......\)

Ta được PT : \(\dfrac{1-a}{a^2}+\dfrac{1-b}{b^2}+\dfrac{1-c}{c^2}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\dfrac{1}{a^2}-\dfrac{1}{a}+\dfrac{1}{4}+\dfrac{1}{b^2}-\dfrac{1}{b}+\dfrac{1}{4}+\dfrac{1}{c^2}-\dfrac{1}{c}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2=0\)

- Thấy : \(\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2\ge0,......\)

\(\Rightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2\ge0\)

- Dấu " = " xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{2}\\\dfrac{1}{b}=\dfrac{1}{2}\\\dfrac{1}{c}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)

- Thay lại a. b. c ta được : \(\left\{{}\begin{matrix}\sqrt{x-2019}=2\\\sqrt{y-2020}=2\\\sqrt{z-2021}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2019=4\\y-2020=4\\z-2021=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2023\\y=2024\\z=2025\end{matrix}\right.\) ( TM )

Vậy ...

xét x=y,x>y và x<y chú ý tới điều kiện x,y thuộc -1;1 nữa 

Thử nhé

Vì P là bất đẳng thức đối xứng nên dự đoán điểm rơi \(x=y=z=\dfrac{\sqrt{2021}}{3}\)

Thay vo P ta duoc \(P=4.\sqrt{2021}\)

----------------------------------------------------------

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

Cauchy-Schwarz:

\(\Rightarrow\left(y+z\right)\left(z+x\right)\ge\left(z+\sqrt{xy}\right)^2\Rightarrow\sqrt{\left(y+z\right)\left(z+x\right)}\ge z+\sqrt{xy}\)

\(\Rightarrow P\ge\sum\dfrac{\left(x+y\right)\left(z+\sqrt{xy}\right)}{z}\ge\sum\dfrac{xz+yz+x\sqrt{y}+y\sqrt{x}}{z}=\sum x+y+\dfrac{\left(x+y\right)\sqrt{xy}}{z}\ge\sum x+y+\dfrac{2xy}{z}\)

\(\Rightarrow P\ge2(x+y+z)+2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\)

Cauchy-Schwarz: \(\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\ge\left(\sqrt{\dfrac{xy}{z}.\dfrac{yz}{z}}+\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\right)^2=\left(x+y+z\right)^2\)

\(\Rightarrow P\ge2(x+y+z)+2\left(x+y+z\right)=4\left(x+y+z\right)=4\sqrt{2021}\)

\("="\Leftrightarrow x=y=z=\dfrac{\sqrt{2021}}{3}\)

Lời giải:

\((\sqrt{x}+\sqrt{y}+\sqrt{z})^2=5^2=25\)

\(\Rightarrow x+y+z+2(\sqrt{xy}+\sqrt{yz}+\sqrt{xz})=25\Rightarrow \sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\frac{25-9}{2}=8\)

\(\Rightarrow xy+yz+xz+2\sqrt{xyz}(\sqrt{x}+\sqrt{y}+\sqrt{z})=64\)

\(\Rightarrow xy+yz+xz+10\sqrt{xyz}=64\)

Thay vào PT(3):

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}\Rightarrow \frac{xy+yz+xz}{xy}=\frac{3}{2}\)

\(\Rightarrow \frac{64-10\sqrt{xyz}}{xyz}=\frac{3}{2}\)

Đặt \(\sqrt{xyz}=t\Rightarrow \frac{64-10t}{t^2}=\frac{3}{2}\Rightarrow 3t^2+20t-128=0\)

\(\Rightarrow t=4\) (chọn) hoặc \(t=-\frac{32}{3}< 0\) (loại)

\(\Rightarrow \sqrt{xy}=\frac{4}{\sqrt{z}}\)

\(\Rightarrow 8=\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\frac{4}{\sqrt{z}}+\sqrt{z}(\sqrt{x}+\sqrt{y})=\frac{4}{\sqrt{z}}+\sqrt{z}(5-\sqrt{z})\)

Đặt \(\sqrt{z}=k\Rightarrow 8k=4+5k^2-k^3\)

\(\Rightarrow k^3-5k^2+8k-4=0\)

\(\Rightarrow k^2(k-1)-4(k^2-2k+1)=0\)

\(\Rightarrow (k-1)(k-2)^2=0\Rightarrow k=1; k=2\)

Nếu $k=1$ suy ra $z=1$. Thay vào giải hpt 2 ẩn ta thu được $x=y=4$

Nếu $k=2$ thì $z=4$. Thay vào giải hpt 2 ẩn ta thu được $(x,y)=(4,1)$ và hoán vị

Vậy $(x,y,z)=(4,4,1)$ và hoán vị của nó.

Đề thiếu. Bạn coi lại đề

Đề thiếu rồi bạn

ĐKXĐ: \(x;y;z\ge0\)

Đặt \(\left(\dfrac{\sqrt{x}}{5};\dfrac{\sqrt{y}}{4};\dfrac{\sqrt{z}}{3}\right)=\left(a;b;c\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}5a+4b+3c=12\\10a+20b+30c=60abc\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5a+4b+3c=12\\a+2b+3c=6abc\end{matrix}\right.\)

Ta có:

\(12=\left(a+a+a+a+a\right)+\left(b+b+b+b\right)+\left(c+c+c\right)\ge12\sqrt[12]{a^5b^4c^3}\)

\(\Rightarrow a^5b^4c^3\le1\) (1)

\(6abc=a+b+b+c+c+c\ge6\sqrt[6]{ab^2c^3}\)

\(\Rightarrow a^6b^6c^6\ge ab^2c^3\Rightarrow a^5b^4c^3\ge1\) (2)

(1);(2) \(\Rightarrow a^5b^4c^3=1\)

Đẳng thức xảy ra khi và chỉ khi \(a=b=c=1\)

\(\Rightarrow\left(x;y;z\right)=\left(25;16;9\right)\)