\(4=2^x+2^y\ge2\sqrt{2^{x+y}}\Rightarrow2^{x+y}\le4\Rightarrow x+y\le2\)

\(\Rightarrow xy\le1\)

\(P=4x^2y^2+2x^3+2y^3+10xy\)

\(P=4x^2y^2+10xy+2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)

\(P\le4x^2y^2+10xy+4\left(4-3xy\right)=4x^2y^2-2xy+16\)

Đặt \(xy=t\Rightarrow0< t\le1\)

Xét hàm \(f\left(t\right)=4t^2-2t+16\) trên \((0;1]\)

\(\Rightarrow...\)

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

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

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

Dấu "=" xảy ra tai x=y=1/2

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

Ta có : \(\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)\ge4xy\)

và \(\left(1-x\right)\left(1-y\right)=1-\left(x+y\right)+xy\le1-2\sqrt{xy}+xy\)

\(\Rightarrow1-2\sqrt{xy}+xy\ge4xy\Leftrightarrow0\) <\(xy\le\frac{1}{9}\)

Dễ chứng minh : \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\le\frac{1}{1+xy};\left(x,y\in\left(0;1\right)\right)\)

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

\(3xy-\left(x^2+y^2\right)=xy-\left(x-y\right)^2\le xy\)

\(\Rightarrow P\le\frac{2}{\sqrt{1+xy}}+xy=\frac{2}{\sqrt{1+t}}+t\), \(\left(t=xy\right)\), (0<\(t\le\frac{1}{9}\)

Xét hàm số :

\(f\left(t\right)=\frac{2}{\sqrt{t+1}}+t\) ,  (0<\(t\le\frac{1}{9}\)

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

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)

\(\Rightarrow b\left(b^2+1\right)-3a^2=\left(a^2+1\right)a-3b^2\)

\(\Rightarrow a^3-b^3+3a^2-3b^2+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)\left(3a+3b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+3a+3b+1\right)=0\)

\(\Leftrightarrow a=b\Rightarrow\sqrt{2x+3}=\sqrt{y}\)

\(\Rightarrow y=2x+3\)

\(\Rightarrow M=x\left(2x+3\right)+3\left(2x+3\right)-4x^2-3\) tới đây chắc chỉ cần bấm máy