ĐK: \(x\ge-1;y\ge3;z\ge1\)

\(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}=\dfrac{1}{2}\left(x+y+z\right)\)

\(\Leftrightarrow x+1-2\sqrt{x+1}+1+y-3-2\sqrt{y-3}+1+z-1-2\sqrt{z-1}+1=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2=0\)

Ta thấy: \(\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2\ge0\)

Đẳng thức xảy ra khi:

\(\left\{{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{y-3}=1\\\sqrt{z-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)

Cách khác:

ĐK: \(x\ge-1;y\ge3;z\ge1\)

Áp dụng BĐT \(ab\le\dfrac{a^2+b^2}{2}\).

\(\sqrt{x+1}\le\dfrac{x+1+1}{2}=\dfrac{x+2}{2}\)

\(\sqrt{y-3}\le\dfrac{y-3+1}{2}=\dfrac{y-2}{2}\)

\(\sqrt{z-1}\le\dfrac{z-1+1}{2}=\dfrac{z}{2}\)

Cộng vế theo vế các BĐT trên ta được:

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

Đẳng thức xảy ra khi:

\(\left\{{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{y-3}=1\\\sqrt{z-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)

ĐKXĐ: \(x\ge-1;y\ge3;z\ge1\)

\(\Leftrightarrow x+y+z-2\sqrt{x+1}-2\sqrt{y-3}-2\sqrt{z-1}=0\)

\(\Leftrightarrow\left(x+1-2\sqrt{x+1}+1\right)+\left(y-3-2\sqrt{y-3}+1\right)+\left(z-1-2\sqrt{z-1}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}-1=0\\\sqrt{y-3}-1=0\\\sqrt{z-1}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)

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:

Ta có:

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

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

Áp dụng BĐT Bunhiacopxky:

\((x+y)(x+z)\geq (x+\sqrt{yz})^2\) và tương tự với những biểu thức khác suy ra:

\(A\geq \frac{(y+z)(x+\sqrt{yz})}{x}+\frac{(z+x)(y+\sqrt{xz})}{y}+\frac{(x+y)(z+\sqrt{xy})}{z}\)

hay \(A\geq 2(x+y+z)+\frac{(y+z)\sqrt{yz}}{x}+\frac{(z+x)\sqrt{zx}}{y}+\frac{(x+y)\sqrt{xy}}{z}\)

hay \(A\geq 2(x+y+z)+\underbrace{\frac{yz(y+z)\sqrt{yz}+xz(x+z)\sqrt{xz}+xy(x+y)\sqrt{xy}}{xyz}}_{M}\)

Đặt \((x,y,z)=(a^2,b^2,c^2)\)

Khi đó: \(M=\frac{a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(a^2+c^2)}{a^2b^2c^2}\)

Áp dụng BĐT AM-GM:

\(a^5b^3+a^3b^5\geq 2\sqrt{a^8b^8}=2a^4b^4\)

\(b^5c^3+c^5b^3\geq 2b^4c^4\)

\(c^5a^3+a^5c^3\geq 2c^4a^4\)

\(\Rightarrow a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(c^2+a^2)\geq 2(a^4b^4+b^4c^4+c^4a^4)\) (1)

(cộng các BĐT theo vế)

Tiếp tục AM-GM:

\(a^4b^4+b^4c^4\geq 2a^2b^4c^2; b^4c^4+c^4a^4\geq 2a^2b^2c^4; c^4a^4+a^4b^4\geq 2a^4b^2c^2\)

\(\Rightarrow a^4b^4+b^4c^4+c^4a^4\geq a^2b^2c^2(a^2+b^2+c^2)\) (2)

Từ\((1); (2)\Rightarrow a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(c^2+a^2)\geq 2a^2b^2c^2(a^2+b^2+c^2)\)

\(\Rightarrow M\geq 2(a^2+b^2+c^2)=2(x+y+z)\)

Do đó: \(A\geq 2(x+y+z)+M\geq 4(x+y+z)\Leftrightarrow A\geq 4\sqrt{2}\)

Vậy \(A_{\min}=4\sqrt{2}\Leftrightarrow x=y=z=\frac{\sqrt{2}}{3}\)

Lời giải:

Ta có:

A=√(x+y)(y+z)(z+x)(√y+zx+√z+xy+√x+yz)

A=(y+z)√(x+y)(x+z)x+(z+x)√(y+z)(y+x)y+(x+y)√(z+x)(z+y)z

Áp dụng BĐT Bunhiacopxky:

(x+y)(x+z)≥(x+√yz)2 và tương tự với những biểu thức khác suy ra:

A≥(y+z)(x+√yz)x+(z+x)(y+√xz)y+(x+y)(z+√xy)z

hay A≥2(x+y+z)+(y+z)√yzx+(z+x)√zxy+(x+y)√xyz

hay A≥2(x+y+z)+yz(y+z)√yz+xz(x+z)√xz+xy(x+y)√xyxyz M 

Đặt (x,y,z)=(a2,b2,c2)

Khi đó: M=a3b3(a2+b2)+b3c3(b2+c2)+c3a3(a2+c2)a2b2c2

Áp dụng BĐT AM-GM:

a5b3+a3b5≥2√a8b8=2a4b4

b5c3+c5b3≥2b4c4

c5a3+a5c3≥2c4a4

⇒a3b3(a2+b2)+b3c3(b2+c2)+c3a3(c2+a2)≥2(a4b4+b4c4+c4a4) (1)

(cộng các BĐT theo vế)

Tiếp tục AM-GM:

a4b4+b4c4≥2a2b4c2;b4c4+c4a4≥2a2b2c4;c4a4+a4b4≥2a4b2c2 

⇒a4b4+b4c4+c4a4≥a2b2c2(a2+b2+c2) (2)

Từ(1);(2)⇒a3b3(a2+b2)+b3c3(b2+c2)+c3a3(c2+a2)≥2a2b2c2(a2+b2+c2)

⇒M≥2(a2+b2+c2)=2(x+y+z)

Do đó: A≥2(x+y+z)+M≥4(x+y+z)⇔A≥4√2

Vậy Amin=4√2⇔x=y=z=√23

Đề bài sai, biểu thức này ko có min

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

\(A_{min}=\dfrac{1}{2}\) khi \(x=y=z=\dfrac{1}{3}\)