\(M^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2xy}{\sqrt{yz}}+\frac{2yz}{\sqrt{zx}}+\frac{2xz}{\sqrt{yz}}=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2x\sqrt{y}}{\sqrt{z}}+\frac{2y\sqrt{z}}{\sqrt{x}}+\frac{2z\sqrt{x}}{\sqrt{y}}\)

Áp dụng bđt Cô-si: \(\frac{x^2}{y}+\frac{x\sqrt{y}}{\sqrt{z}}+\frac{x\sqrt{y}}{\sqrt{z}}+z\ge4\sqrt[4]{\frac{x^2}{y}.\frac{x\sqrt{y}}{\sqrt{z}}.\frac{x\sqrt{y}}{\sqrt{z}}.z}=4x\)

tương tự \(\frac{y^2}{z}+\frac{y\sqrt{z}}{\sqrt{x}}+\frac{y\sqrt{z}}{\sqrt{x}}+x\ge4y\);\(\frac{z^2}{x}+\frac{z\sqrt{x}}{\sqrt{y}}+\frac{z\sqrt{x}}{\sqrt{y}}+y\ge4z\)

=>\(M^2+x+y+z\ge4\left(x+y+z\right)\Rightarrow M^2\ge3\left(x+y+z\right)\ge3.12=36\Rightarrow M\ge6\)

Dấu "=" xảy ra khi x=y=z=4

Vậy minM=6 khi x=y=z=4

b1: Áp dụng bđt Cauchy Schwarz dạng Engel ta được:

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

=>minP=1 <=> x=y=z=2/3

(x^2-2+1/x^2 ) +( y^2-2+1/y^2) +(z^2-2+1/z^2) =0

=> (x-1/x)^2 +(y-1/y)^2+(z-1/z)^2=0

suy ra x-1/x=0 

          y-1/y=0

         z-1/z=0

.....

Ta có: \(x^2+\frac{1}{x^2}\ge2\sqrt{x^2.\frac{1}{x^2}}=2\)

\(y^2+\frac{1}{y^2}\ge2\sqrt{y^2.\frac{1}{y^2}}=2\)

\(z^2+\frac{1}{z^2}\ge2\sqrt{x^2.\frac{1}{z^2}}=2\)

\(\Rightarrow VT\ge6\)

Dấu "=" khi \(\orbr{\begin{cases}x=y=z=1\\x=y=z=-1\end{cases}}\)

Bài này thì chắc cô si ngược dấu thôi:v

\(LHS=\Sigma\frac{x}{1+y^2}=\Sigma x.\left(1-\frac{y^2}{1+y^2}\right)\)

\(\ge\Sigma x\left(1-\frac{y}{2}\right)=x+y+z-\frac{xy+yz+zx}{2}\)

\(\ge x+y+z-\frac{\left(x+y+z\right)^2}{6}=\frac{3}{2}\)

P/s: check xem có ngược dấu chỗ nào ko:v

Dự đoán khi \(x=y=z=\sqrt{3}\) vậy dc GTNN là \(\frac{3\sqrt{3}}{2}\), cần c/m: \(P\ge\frac{3\sqrt{3}}{2}\)

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

\(\LeftrightarrowΣ\frac{y^3z^3}{y^2+z^2}\ge\frac{3}{2}\sqrt{\frac{3x^4y^4z^4}{x^2y^2+x^2z^2+y^2z^2}}\).Đặt \(\hept{\begin{cases}yz=a\\xz=b\\xy=c\end{cases}}\)

Khi đó ta cần chứng minh \(Σ\frac{a^3}{\frac{ac}{b}+\frac{ab}{c}}\ge\frac{3}{2}\sqrt{\frac{3a^2b^2c^2}{a^2+b^2+c^2}}\)

\(\LeftrightarrowΣ\frac{a^2}{b^2+c^2}\ge\frac{3}{2}\sqrt{\frac{3}{a^2+b^2+c^2}}\) và từ BĐT thuần nhất cuối , ta có thế khẳng định rằng \(a^2+b^2+c^2=3\)

Có nghĩa là ta cần c/m \(Σ\frac{a}{3-a^2}\ge\frac{3}{2}\LeftrightarrowΣ\left(\frac{a}{3-a^2}-\frac{1}{2}\right)\ge0\)

\(\LeftrightarrowΣ\frac{\left(a-1\right)\left(a+3\right)}{3-a^2}\ge0\)\(\LeftrightarrowΣ\left(\frac{\left(a-1\right)\left(a+3\right)}{3-a^2}-\left(a^2-1\right)\right)\ge0\)

\(\LeftrightarrowΣ\frac{a\left(a+2\right)\left(a-1\right)^2}{3-a^2}\ge0\) . XOng!

a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)

Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2

b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)

Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)

Đặt \(P=\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Do x,y,z là các số thực dương nên ta biến đổi \(P=\frac{1}{\sqrt{1+\frac{1}{x^2}}}+\frac{1}{\sqrt{1+\frac{1}{y^2}}}+\frac{1}{\sqrt{1+\frac{1}{z^2}}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Đặt \(a=\frac{1}{x^2};b=\frac{1}{y^2};c=\frac{1}{z^2}\left(a,b,c>0\right)\)thì \(xy+yz+zx=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}=1\)và \(P=\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}+a+b+c\)

Biến đổi biểu thức P=\(\left(\frac{1}{2\sqrt{a+1}}+\frac{1}{2\sqrt{a+1}}+\frac{a+1}{16}\right)+\left(\frac{1}{2\sqrt{b+1}}+\frac{1}{2\sqrt{b+1}}+\frac{b+1}{16}\right)\)\(+\left(\frac{1}{2\sqrt{c+1}}+\frac{1}{2\sqrt{c+1}}+\frac{c+1}{16}\right)+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{b}-\frac{3}{16}\)

Áp dụng Bất Đẳng Thức Cauchy ta có

\(P\ge3\sqrt[3]{\frac{a+1}{64\left(a+1\right)}}+3\sqrt[3]{\frac{b+1}{64\left(b+1\right)}}+3\sqrt[3]{\frac{c+1}{64\left(c+1\right)}}+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{16}-\frac{3}{16}\)

\(=\frac{33}{16}+\frac{15}{16}\left(a+b+c\right)\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{abc}\)

Mặt khác ta có \(1=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ge3\sqrt[3]{\frac{1}{abc}}\Leftrightarrow abc\ge27\)

\(\Rightarrow P\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{27}=\frac{33}{16}+\frac{15}{16}\cdot9=\frac{21}{2}\)

Dấu "=" xảy ra khi a=b=c hay \(x=y=z=\frac{\sqrt{3}}{3}\)

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

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

Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a;b;c\right)\) thì \(a^2+b^2+c^2=1\) Ta cần chứng minh:

\(P=\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\)

\(=\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\)

\(=\frac{a^2}{a\left(1-a^2\right)}+\frac{b^2}{b\left(1-b^2\right)}+\frac{c^2}{c\left(1-c^2\right)}\)

Theo đánh giá bởi AM - GM ta có:

\(a^2\left(1-a^2\right)^2=\frac{1}{2}\cdot2a^2\cdot\left(1-a^2\right)\left(1-a^2\right)\)

\(\le\frac{1}{2}\left(\frac{2a^2+1-a^2+1-a^2}{3}\right)^3=\frac{4}{27}\)

\(\Rightarrow a\left(1-a^2\right)^2\le\frac{2}{3\sqrt{3}}\Leftrightarrow\frac{a^2}{a\left(1-a\right)^2}\ge\frac{3\sqrt{3}}{2}a^2\)

Tương tự rồi cộng lại ta có ngay điều phải chứng minh