\(x\left(x-z\right)+y\left(y-z\right)=0\)\(\Leftrightarrow\)\(x^2+y^2=z\left(x+y\right)\)

\(\frac{x^3}{z^2+x^2}=x-\frac{z^2x}{z^2+x^2}\ge x-\frac{z^2x}{2zx}=x-\frac{z}{2}\)

\(\frac{y^3}{y^2+z^2}=y-\frac{yz^2}{y^2+z^2}\ge y-\frac{yz^2}{2yz}=y-\frac{z}{2}\)

\(\frac{x^2+y^2+4}{x+y}=\frac{z\left(x+y\right)+4}{x+y}=z-x-y+\frac{4}{x+y}+x+y\ge z-x-y+4\)

Cộng lại ra minP=4, dấu "=" xảy ra khi \(x=y=z=1\)

Do \(x^2+y^2+z^2=1\Rightarrow x^2< 1\Rightarrow x< 1\)

\(\Rightarrow x^5< x^2\)

Tương tự ta có: \(y< 1\Rightarrow y^6< y^2\); \(z< 1\Rightarrow z^7< z^2\)

\(\Rightarrow x^5+y^6+z^7< x^2+y^2+z^2\)

\(\Rightarrow x^5+y^6+z^7< 1\)

Lời giải:
Áp dụng BĐT Cô-si:

$x^3+1+1\geq 3x$

$y^3+1+1\geq 3y$

$z^3+1+1\geq 3z$

$\Rightarrow x^3+y^3+z^3+6\geq 3(x+y+z)\geq 3.3=9$

$\Rightarrow A=x^3+y^3+z^3\geq 3$ 

Vậy $A_{\min}=3$. Giá trị này đạt tại $x=y=z=1$

\(A=\left(x^3+1+1\right)+\left(y^3+1+1\right)+\left(z^3+1+1\right)-6\)

\(A\ge3\sqrt[3]{x^3}+3\sqrt[3]{y^3}+3\sqrt[3]{z^3}-6=3\left(x+y+z\right)-6\ge3.3-6=3\)

\(A_{min}=3\) khi \(x=y=z=1\)

Ta cần chứng minh:\(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\sqrt{3}\)

Áp dụng bất đẳng thức Bunhiacopxki, ta được:

\(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\dfrac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\)

Mặt khác, ta có:

\(\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(\left(x+y+xy\right)+\left(y+z+yz\right)+\left(z+x+zx\right)\right)\)

\(\Leftrightarrow\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(6+xy+yz+zx\right)\)Lại có:

\(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{9}{3}=3\)

\(\Rightarrow\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(6+3\right)=27\)

\(\Rightarrow\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\le3\sqrt{3}\)

\(\Rightarrow\dfrac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\ge\dfrac{9}{3\sqrt{3}}=\sqrt{3}\)

Do đó \(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\sqrt{3}\)

Dấu bằng xảy ra \(\Leftrightarrow x=y=z=1\).

\(x,y,z>0\)

Áp dụng BĐT Caushy cho 3 số ta có:

\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)

\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)

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

Áp dụng BĐT Caushy-Schwarz ta có:

\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)

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

\(P=0\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=0\)

hiiiii

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