Ta có : \(3\sqrt{xyz}=\sqrt{x}^2+\sqrt{y}^3+\sqrt{z}^3\ge3\sqrt[3]{\sqrt{x}^3\sqrt{y}^3\sqrt{z}^3}=3\sqrt{x}\sqrt{y}\sqrt{z}=3\sqrt{xyz}.\)

Dấu = xảy ra

=> x =y =z

=> A = (1+1)(1+1)(1+1) =8

[Verse 1: Rap Monster]
Ai tai
Kotoba ni suru to yokei
Ai tai
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Tada sugi te iku
Toki no naka mizukara o seme mata
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[Verse 2: Rap Monster, Jungkook]
Marude fuyu no yō sa natsu de mo fubuku yō sa
Kokoro o nose ta ressha sudeni yuki no naka
Isso kimi to chikyū no uragawa e te o tsukan de mō nige tai
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[Verse 3: Jimin, V]
Kono ōzora o mau koyuki no yō
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[Pre-Chorus: Jungkook, J-Hope, V, Jin]
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[Chorus: Jungkook, Jin, J-Hope]
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[Chorus: Jungkook, Jin, J-Hope]
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\(\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)

\(\Rightarrow\dfrac{x}{x+\sqrt{x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

\(\dfrac{z}{z+\sqrt{z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế:

\(VT\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

Áp dụng bất đẳng thức cô si ta có :

\(x\sqrt{x}+y\sqrt{y}+z\sqrt{z}=\left(\sqrt{x}\right)^3+\left(\sqrt{y}\right)^3+\left(\sqrt{z}\right)^3\ge3\sqrt[3]{\left(\sqrt{xyz}\right)^3}=3\sqrt{xyz}\)Dấu "=" xảy ra khi :\(\sqrt{x}=\sqrt{y}=\sqrt{z}\)

Do đó :\(A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

Vậy A=8

Đề bài chắc chắn là có vấn đề

Thử với \(x=y=z=\dfrac{1}{3}\) thì \(VT=\dfrac{\sqrt{2}}{4}< 2\)

Như bạn sửa điều kiện thành \(x^3+y^3+z^3=1\) thì dấu "=" không xảy ra

Việc chứng minh vế trái lớn hơn 2 (một cách tuyệt đối) khá đơn giản:

\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\ge\dfrac{x^3}{\dfrac{x^2+1-x^2}{2}}=2x^3\)

Làm tương tự với 2 số hạng còn lại, sau đó cộng vế

Nhưng đẳng thức không xảy ra.

Ta có :VT-VP=

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

\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}+\dfrac{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}+\sqrt{z}\right)}{\sqrt{y}+\sqrt{z}}+\dfrac{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{z}+\sqrt{x}\right)}{\sqrt{x}+\sqrt{x}}\)\(=\left(\sqrt{x}-\sqrt{y}\right)+\left(\sqrt{y}-\sqrt{z}\right)+\left(\sqrt{z}-\sqrt{x}\right)=0\)

\(\Rightarrow VT=VP\)

Vậy ...

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\dfrac{1}{\sqrt{x}+2\sqrt{y}}\le\dfrac{1}{9}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{y}}\right)\)

Tương tự cho 2 BĐT trên ta có:

\(\dfrac{1}{3}VP\le\dfrac{1}{9}\cdot3\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)\)

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

Xảy ra khi \(x=y=z\)

Lời giải:
\(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{x+y-z}\Leftrightarrow \frac{x+y}{xy}=\frac{1}{z}+\frac{1}{x+y-z}=\frac{x+y}{z(x+y-z)}\)

\(\Leftrightarrow (x+y)(\frac{1}{xy}-\frac{1}{z(x+y-z)})=0\)

\(\Leftrightarrow (x+y).\frac{z(x+y-z)-xy}{xyz(x+y-z)}=0\)

\(\Leftrightarrow (x+y).\frac{(z-x)(y-z)}{xyz(x+y-z)}=0\)

\(\Leftrightarrow (x+y)(z-x)(y-z)=0\)

Xét các TH sau:

TH1: $x+y=0$. TH này loại do ĐKXĐ $x,y>0$
TH2: $z-x=0\Leftrightarrow z=x$

$\Leftrightarrow \frac{1}{y}=\frac{2020}{2021}$

\(M=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{y}}=\frac{2}{\sqrt{y}}=2\sqrt{\frac{2020}{2021}}\)

TH3: $y-z=0$ tương tự TH2, ta có \(M=2\sqrt{\frac{2020}{2021}}\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\)

\(\Rightarrow\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}\ge\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\)

\(\Rightarrow\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}-\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\ge0\)

\(\Rightarrow\dfrac{1}{x}-\dfrac{2}{\sqrt{xy}}+\dfrac{1}{y}+\dfrac{1}{y}-\dfrac{2}{\sqrt{yz}}+\dfrac{1}{z}+\dfrac{1}{z}-\dfrac{2}{\sqrt{zx}}+\dfrac{1}{x}\ge0\)

\(\Rightarrow\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{y}}\right)^2+\left(\dfrac{1}{\sqrt{y}}-\dfrac{1}{\sqrt{z}}\right)^2+\left(\dfrac{1}{\sqrt{z}}-\dfrac{1}{\sqrt{x}}\right)^2\ge0\) (luôn đúng)

Dấu = xảy ra khi \(x=y=z\)