Ta có : \(A^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2x\sqrt{y}}{\sqrt{z}}+\frac{2y\sqrt{z}}{\sqrt{x}}+\frac{2z\sqrt{x}}{\sqrt{y}}\)

Áp dụng BĐT Cô-si cho 4 số dương,ta có ;

\(\frac{x^2}{y}+\frac{x\sqrt{y}}{\sqrt{z}}+\frac{x\sqrt{y}}{\sqrt{z}}+z\ge4\sqrt[4]{\frac{x^2.x^2.y.z}{yz}}=4x\)

Tương tự : ....

\(\Rightarrow A^2\ge4\left(x+y+z\right)-\left(x+y+z\right)=3\left(x+y+z\right)\ge36\)

\(\Rightarrow A\ge6\)

Dấu "=" xảy ra khi x = y = z = 4

Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)\rightarrow\left(a;b;c\right)\)

Khi đó \(a^2+b^2+c^2\ge12\) ta cần tìm GTNN của  \(A=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)

\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge2\sqrt{\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\left(a+b+c\right)}\)

Ta có:\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2b+b^2c+c^2a}\)

Mà \(\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge3\left(a^2+b^2+c^2\right)\) ( cơ bản )

\(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge2\sqrt{3\left(a^2+b^2+c^2\right)}=12\)

\(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge12-\left(a+b+c\right)\)

Chứng minh được \(a+b+c\le6\) là OKE nhưng có vẻ không ổn lắm :))

b, Ta có 

\(\frac{\sqrt{x}+1}{y+1}=\frac{\left(\sqrt{x}+1\right)\left(y+1\right)-y-y\sqrt{x}}{y+1}=\sqrt{x}+1-\frac{y\left(\sqrt{x}+1\right)}{y+1}\)

Mà \(y+1\ge2\sqrt{y}\)

=> \(\frac{\sqrt{x}+1}{y+1}\ge\sqrt{x}+1-\frac{1}{2}\sqrt{y}\left(\sqrt{x}+1\right)\)

Khi đó

\(P\ge\frac{1}{2}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3-\frac{1}{2}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)\)

Mà \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}=3\)

=> \(P\ge\frac{1}{2}.3+3-\frac{3}{2}=3\)

Vậy MinP=3 khi x=y=z=1

của ? 

=> THIẾU ĐỀ

Ta có: P² = \(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}+2\left[\dfrac{xy}{\sqrt{yz}}+\dfrac{yz}{\sqrt{zx}}+\dfrac{zx}{\sqrt{xy}}\right]\)

Áp dụng BĐT Cô-si ta có: \(\dfrac{x^2}{y}+\dfrac{xy}{\sqrt{yz}}\dfrac{xy}{\sqrt{yz}}+z\ge4x\)

Tương tự cộng lại ta có: \(P^2+\left(x+y+z\right)\ge4\left(x+y+z\right)\)

\(\Rightarrow P^2\ge3\left(x+y+z\right)\ge36\Rightarrow P\ge6\)

Dấu"=" xảy ra \(\Leftrightarrow\)x=y=z=4

\(A=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\ge\frac{2}{x+1}+\frac{2}{y+1}+\frac{2}{z+1}\ge\frac{18}{x+y+z+3}=3\)

cảm ơn nha

=3 ban nhe.kn voi minh nha

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

\(P\ge\frac{\left(x+y+z\right)^2}{\frac{x\left(y+4\right)}{4}+\frac{y\left(z+4\right)}{4}+\frac{z\left(x+4\right)}{4}}=\frac{4\left(x+y+z\right)^2}{xy+yz+zx+4\left(x+y+z\right)}\)

\(P\ge\frac{4\left(x+y+z\right)^2}{\frac{\left(x+y+z\right)^2}{3}+4\left(x+y+z\right)}=\frac{12\left(x+y+z\right)}{x+y+z+12}=12-\frac{144}{x+y+z+12}\ge12-\frac{144}{12+12}=6\)

\(\Rightarrow P_{min}=6\) khi \(x=y=z=4\)