Ta có \(2=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\Leftrightarrow xy\ge1\)

\(A=\dfrac{1}{x^4+y^2+2xy^2}+\dfrac{1}{x^2+y^4+2x^2y}\\ \le\dfrac{1}{4\sqrt[4]{x^6y^6}}+\dfrac{1}{4\sqrt[4]{x^6y^6}}=\dfrac{1}{4xy}+\dfrac{1}{4xy}\\ \le\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)

Dấu \("="\Leftrightarrow x=y=1\)

1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có 

\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)

\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)

    \(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)

\(P=\dfrac{x}{\sqrt{2.\dfrac{1}{2}x+2yz}}+\dfrac{y}{\sqrt{2.\dfrac{1}{2}y+zx}}+\dfrac{z}{\sqrt{2.\dfrac{1}{2}z+xy}}\)

\(=\dfrac{x}{\sqrt{2x\left(x+y+z\right)+yz}}+\dfrac{y}{\sqrt{2y\left(x+y+z\right)+2zx}}+\dfrac{z}{\sqrt{2z\left(x+y+z\right)+2xy}}\)

\(=\dfrac{x}{\sqrt{2\left(x+y\right)\left(x+z\right)}}+\dfrac{y}{\sqrt{2\left(x+y\right)\left(y+z\right)}}+\dfrac{z}{\sqrt{2\left(x+z\right)\left(y+z\right)}}\)

\(=\dfrac{1}{2\sqrt{2}}.2\sqrt{\dfrac{x}{x+y}}.\sqrt{\dfrac{x}{x+z}}+\dfrac{1}{2\sqrt{2}}.2\sqrt{\dfrac{y}{x+y}}.\sqrt{\dfrac{y}{y+z}}+\dfrac{1}{2\sqrt{2}}.2\sqrt{\dfrac{z}{x+z}}.\sqrt{\dfrac{z}{y+z}}\)

\(\le\dfrac{1}{2\sqrt{2}}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)\)

\(=\dfrac{3}{2\sqrt{2}}\)

Dấu "=" xảy ra tại  \(x=y=z=\dfrac{1}{6}\)

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\)

Đặt \(\dfrac{x}{y}=a\Rightarrow0< a\le\dfrac{1}{4}\)

\(P=\dfrac{\left(\dfrac{x}{y}\right)^2-\dfrac{2x}{y}+2}{\dfrac{x}{y}+1}=\dfrac{a^2-2a+2}{a+1}=\dfrac{4a^2-8a+8}{4\left(a+1\right)}=\dfrac{4a^2-13a+3+5\left(a+1\right)}{4\left(a+1\right)}\)

\(P=\dfrac{5}{4}+\dfrac{\left(1-4a\right)\left(3-a\right)}{4\left(a+1\right)}\ge\dfrac{5}{4}\)

Dấu "=" xảy ra khi \(a=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

Do \(1\le x\le2\Rightarrow\left(x-1\right)\left(x-2\right)\le0\)

\(\Leftrightarrow x^2+2\le3x\)

Hoàn toàn tương tự ta có \(y^2+2\le3y\)

Do đó: \(P\ge\dfrac{x+2y}{3x+3y+3}+\dfrac{2x+y}{3x+3y+3}+\dfrac{1}{4\left(x+y-1\right)}\)

\(P\ge\dfrac{x+y}{x+y+1}+\dfrac{1}{4\left(x+y-1\right)}\)

Đặt \(a=x+y-1\Rightarrow1\le a\le3\)

\(\Rightarrow P\ge f\left(a\right)=\dfrac{a+1}{a+2}+\dfrac{1}{4a}\)

\(f'\left(a\right)=\dfrac{3a^2-4a-4}{4a^2\left(a+2\right)^2}=\dfrac{\left(a-2\right)\left(3a+2\right)}{4a^2\left(a+2\right)^2}=0\Rightarrow a=2\)

\(f\left(1\right)=\dfrac{11}{12}\) ; \(f\left(2\right)=\dfrac{7}{8}\) ; \(f\left(3\right)=\dfrac{53}{60}\)

\(\Rightarrow f\left(a\right)\ge\dfrac{7}{8}\Rightarrow P_{min}=\dfrac{7}{8}\) khi \(\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)

Lời giải:
\(P=\sum \frac{1}{2xy^2+1}=\sum (1-\frac{2xy^2}{2xy^2+1})\)

\(=3-2\sum\frac{xy^2}{2xy^2+1}\geq 3-2\sum \frac{xy^2}{3\sqrt[3]{x^2y^4}}\) theo BĐT AM-GM.

\(=3-\frac{2}{3}\sum \sqrt[3]{xy^2}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt[3]{xy^2}\leq \frac{x+y+y}{3}\Rightarrow \sum \sqrt[3]{xy^2}\leq \frac{3(x+y+z)}{3}=3\)

$\Rightarrow P\geq 3-\frac{2}{3}.3=1$

Vậy $P_{\min}=1$. Giá trị này đạt tại $x=y=z=1$