Vì xyz=1\(\Rightarrow x^2\left(y+z\right)\ge2x^2\sqrt{yz}=2x\sqrt{x}\)

Tương tự \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2=\left(x+y\right)\ge2z\sqrt{z}\)

\(\Rightarrow P\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)

Đặt \(x\sqrt{x}+2y\sqrt{y}=a;y\sqrt{y}+2z\sqrt{z}=b;z\sqrt{z}+2x\sqrt{x}=c\)

\(\Rightarrow x\sqrt{x}=\frac{4c+a-2b}{9};y\sqrt{y}=\frac{4a+b-2c}{9};z\sqrt{z}=\frac{4b+c-2a}{9}\)

\(\Rightarrow P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{a}+\frac{4b+c-2a}{b}\right)\)

\(=\frac{2}{9}\text{ }\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\ge\frac{2}{9}\left(4.3+2-6\right)=2\)

Min P =2 khi và chỉ khi a=b=c khi va chỉ khi x=y=z=1

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

\(\ge\frac{2}{\sqrt{xy}}\sqrt{1+x^2y^2}=2\sqrt{\frac{1}{xy}+xy}=2\sqrt{\frac{1}{16xy}+xy+\frac{15}{16xy}}\)

\(\ge2\sqrt{2\sqrt{\frac{1}{16xy}\cdot xy}+\frac{15}{4\left(x+y\right)^2}}=2\sqrt{\frac{1}{2}+\frac{15}{4}}=\sqrt{17}\)

Dấu "=" xảy ra tai x=y=1/2

Ta có \(\left(2x^2+y^2+3\right)\left(2+1+3\right)\ge\left(2x+y+3\right)^2\)

=> \(\frac{1}{\sqrt{2x^2+y^2+3}}\le\frac{\sqrt{6}}{2x+y+3}\)

Mà \(\frac{1}{2x+y+3}=\frac{1}{x+x+y+1+1+1}\le\frac{1}{36}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+3\right)\)

=> \(\frac{1}{\sqrt{2x^2+y^2+3}}\le\frac{\sqrt{6}}{36}\left(\frac{2}{x}+\frac{1}{y}+3\right)\)

Khi đó 

\(P\le\frac{\sqrt{6}}{36}\left(\frac{3}{x}+\frac{3}{y}+\frac{3}{z}+9\right)=\frac{\sqrt{6}}{36}.18=\frac{\sqrt{6}}{2}\)

Dấu bằng xảy ra khi x=y=z=1

Vậy \(MaxP=\frac{\sqrt{6}}{2}\)khi x=y=z=1

dễ vãi mà ko giải đc NGU

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

\(\ge2\cdot\frac{1}{\sqrt{xy}}\sqrt{1+x^2y^2}\)

\(=2\cdot\sqrt{\frac{1}{xy}+xy}\)

\(=2\cdot\sqrt{xy+\frac{1}{16xy}+\frac{15}{xy}}\)

\(\ge2\sqrt{2\sqrt{xy\cdot\frac{1}{16xy}}+\frac{15}{16xy}}\left(1\right)\)

Áp dụng BĐT phụ \(ab\le\frac{\left(a+b\right)^2}{4}\) ta có:

\(\left(1\right)\ge2\cdot\sqrt{\frac{1}{2}+\frac{15}{4\cdot\left(x+y\right)^2}}\ge2\sqrt{\frac{1}{2}+\frac{15}{4}}=\sqrt{17}\)

Dấu "=" xảy ra tại \(x=y=\frac{1}{2}\)

khó vậy? 

Áp dụng bất đẳng thức Minkowski ta có:

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

\(\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{9}{x+y+z}\right)^2}=\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)

\(=\sqrt{\left[\left(x+y+z\right)^2+\frac{1}{\left(x+y+z\right)^2}\right]+\frac{80}{\left(x+y+z\right)^2}}\)

\(\ge\sqrt{2\sqrt{\left(x+y+z\right)^2\cdot\frac{1}{\left(x+y+z\right)^2}}+\frac{80}{1}}=\sqrt{82}\)

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

Áp dụng bất đẳng thức Minkowski ta có:

√x2+1x2 +√y2+1y2 +√z2+1z2 ≥√(x+y+z)2+(1x +1y +1z )2

≥√(x+y+z)2+(9x+y+z )2=√(x+y+z)2+81(x+y+z)2 

=√[(x+y+z)2+1(x+y+z)2 ]+80(x+y+z)2 

≥√2√(x+y+z)2·1(x+y+z)2 +801 =√82

Dấu "=" xảy ra khi: x=y=z=13 

mình làm ra rồi khỏi cần giúp nữa

ta có: \(\frac{\sqrt{2x^2+y^2}}{xy}=\sqrt{\frac{2}{y^2}+\frac{1}{x^2}}\)

Áp dụng BĐT bunyakovsky:\(\left(2+1\right)\left(\frac{2}{y^2}+\frac{1}{x^2}\right)\ge\left(\frac{2}{y}+\frac{1}{x}\right)^2\)

\(\Rightarrow\frac{2}{y^2}+\frac{1}{x^2}\ge\frac{1}{3}\left(\frac{2}{y}+\frac{1}{x}\right)^2\).....bla bla