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\(P=x^2+y^2+z^2+\dfrac{20}{x+y+z}\ge\dfrac{\left(x+y+z\right)^2}{3}+\dfrac{20}{x+y+z}\)

\(\Leftrightarrow P\ge\dfrac{\left(x+y+z\right)^2}{3}+\dfrac{9}{x+y+z}+\dfrac{9}{x+y+z}+\dfrac{2}{x+y+z}\)

\(\Leftrightarrow P\ge3\sqrt[3]{\dfrac{\left(x+y+z\right)^2}{3}.\dfrac{9}{x+y+z}.\dfrac{9}{x+y+z}}+\dfrac{2}{3}\)

 (theo AM-GM và do \(x+y+z\le3\Rightarrow\dfrac{2}{x+y+z}\ge\dfrac{2}{3}\))

\(\Leftrightarrow P\ge\dfrac{29}{3}\)

Dấu = xảy ra khi x=y=z=1

Vậy minP\(=\dfrac{29}{3}\)

Ta có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow x^2-2xy+y^2+x^2-2xz+z^2+y^2+2yz+z^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2xz+2yz\)

\(\Leftrightarrow3x^2+3y^2+3z^2\ge x^2+y^2+z^2+2xy+2xz+2yz\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge3^2\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge9\)

Áp dụng Bđt Bunhiacopxki cho các cặp số dương \(\left(1;x\right);\left(1;y\right);\left(1;z\right)\) 

\(\left(1.x+1.y+1.z\right)^2\le\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow P=x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{9}{3}=3\)

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

Vậy \(GTNN\left(P\right)=3\left(tạix=y=z=1\right)\)

Ta có: \(2x^3+2y^3-\left(x+y\right)\left(x^2+y^2\right)=\left(x-y\right)^2\left(x+y\right)\ge0\)

\(\Rightarrow\dfrac{x^3+y^3}{x^2+y^2}\ge\dfrac{x+y}{2}\)

Tương tự: \(\dfrac{y^3+z^3}{y^2+z^2}\ge\dfrac{y+z}{2}\) ; \(\dfrac{z^3+x^3}{z^2+x^2}\ge\dfrac{z+x}{2}\)

Cộng vế: \(P\ge x+y+z\ge6\)

\(P_{min}=6\) khi \(x=y=z=2\)

Lời giải:

Áp dụng BĐT AM-GM:

$x^2+\frac{1}{2x}+\frac{1}{2x}\geq 3\sqrt[3]{\frac{1}{4}}$

Tương tự:

$y^2+\frac{1}{2y}+\frac{1}{2y}\geq 3\sqrt[3]{\frac{1}{4}}$

$z^2+\frac{1}{2z}+\frac{1}{2z}\geq 3\sqrt[3]{\frac{1}{4}}$

Cộng theo vế:

$A\geq 9\sqrt[3]{\frac{1}{4}}$ (đây chính là $A_{\min}$)

Dấu "=" xảy ra khi $x=y=z=\sqrt[3]{\frac{1}{2}}$

Áp dụng bất đẳng thức Bunhia dạng phân thức cho 3 số ta có:

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

Dấu "=" xảy ra \(\Leftrightarrow\begin{matrix}\dfrac{x}{y+z}=\dfrac{y}{z+x}=\dfrac{z}{x+y}\\x,y,z>0;x+y+z=2\end{matrix}\)

\(\Leftrightarrow x=y=z=\dfrac{2}{3}\)

Áp dụng BĐT Svac-xơ cho 3 số dương có :

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

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{2}{3}\)

Vậy Min biểu thức cho là 1 khi \(x=y=z=\dfrac{2}{3}\)

:(

\(A=\frac{3+x^2}{y+z}+\frac{3+y^2}{z+x}+\frac{3+z^2}{x+y}\)

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

\(\ge3\cdot\frac{9}{2\left(x+y+z\right)}+\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}\)

\(=\frac{27}{2\cdot3}+\frac{3}{2}=6\)

Đẳng thức xảy ra tại 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 \(P_{min}=0\)