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Lời giải:
Áp dụng BĐT AM-GM:
$2A=2x^2y^2(x^2+y^2)=xy.[2xy(x^2+y^2)]\leq \left(\frac{x+y}{2}\right)^2.\left(\frac{2xy+x^2+y^2}{2}\right)^2$
$\Leftrightarrow 2A\leq \frac{(x+y)^6}{16}=\frac{1}{16}$
$\Rightarrow A\leq \frac{1}{32}$
Vậy $A_{\max}=\frac{1}{32}$. Giá trị này đạt được khi $x=y=\frac{1}{2}$
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :\(2x+yz=\left(x+y+z\right)x+yz=x^2+xy+xz+yz=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow\sqrt{2x+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{\left(x+y\right)+\left(x+z\right)}{2}\)(bất đẳng thức cô si)
Cm tương tự :\(\sqrt{2y+xz}\le\dfrac{\left(y+x\right)+\left(y+z\right)}{2}\)
\(\sqrt{2z+xy}\le\dfrac{\left(z+y\right)+\left(z+x\right)}{2}\)
Do đó :P\(\le\dfrac{4\left(x+y+z\right)}{2}=2\left(x+y+z\right)=2\times2=4\)
Dấu "=" xảy ra khi :x=y=z=\(\dfrac{2}{3}\)
Vây giá trị lớn nhất của P=\(\sqrt{2x+yz}+\sqrt{2y+xz}+\sqrt{2z+xy}\) với x+y+z=2 và x,y,z\(\ge0\) là 4 khi x=y=z=\(\dfrac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5x^2+2xy+2y^2-\left(4x^2+4xy+y^2\right)=\left(x-y\right)^2\ge0\\ \Leftrightarrow5x^2+2xy+2y^2\ge4x^2+4xy+y^2=\left(2x+y\right)^2\)
\(\Leftrightarrow P\le\dfrac{1}{2x+y}+\dfrac{1}{2y+z}+\dfrac{1}{2z+x}=\dfrac{1}{9}\left(\dfrac{9}{x+x+y}+\dfrac{9}{y+y+z}+\dfrac{9}{z+z+x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)
Dấu \("="\Leftrightarrow x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(2\left(2x^2+xy+2y^2\right)=3\left(x^2+y^2\right)+\left(x+y\right)^2\ge\dfrac{3}{2}\left(x+y\right)^2+1\left(x+y\right)^2=\dfrac{5}{2}\left(x+y\right)^2\)
\(\Rightarrow\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
Gợi ý. Dùng cái trên.
![](https://rs.olm.vn/images/avt/0.png?1311)
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
Áp dụng bdt cosi-schwar cho 3 số (\(\left(am+bn+cp\right)^2\le\left(a^2+b^2+c^2\right)\)\(\left(m^2+n^2+p^2\right)\)
với a=x,b=y\(\sqrt{2}\);c=z\(\sqrt{5}\); m=\(\sqrt{11-2y^2},n=\sqrt{3-5z^2}\),\(p=\sqrt{2-x^2}\)
82\(\le\left(x^2+2y^2+5z^2\right)\left(11-2y^2+3-5z^2+1-x^2\right)\) <=>64\(\le P\left(16-P\right)\)
<=>P2-16P+64\(\le0< =>\left(P-8\right)^2\le0\) <=>P=8