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 \(x=y=\dfrac{1}{2}\)

x + y = 1

<=> (x + y)² = 1²

<=> x² + y² + 2xy = 1

<=> x² + y² = 1 - 2xy

Ta có:

\(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)

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

= \(\dfrac{x^4-x-y^4+y}{x^3y^3-y^3-x^3+1}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)

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

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

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

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

\(=-\dfrac{2\left(x-y\right)}{x^2y^2+3}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)

= 0 (đpcm)

\(A=\dfrac{2\left(x^3+y^3\right)}{\left(x^4+y^2\right)\left(x^2+y^4\right)}=2.\dfrac{\left(x^3+y^3\right)}{x^4y^4+x^2y^2+x^6+y^6}\)

\(=2.\dfrac{\left(x^3+y^3\right)}{1+1+x^6+y^6}=2.\dfrac{x^3+y^3}{x^6+y^6+2x^3y^3}=2.\dfrac{x^3+y^3}{\left(x^3+y^3\right)^2}=\dfrac{2}{x^3+y^3}\left(1\right)\)

Áp dụng bất đẳng thức Cauchy ta có:

\(x^3+y^3+1\ge3\sqrt{xy.1}=3\)

\(\Rightarrow x^3+y^3\ge2\Rightarrow\dfrac{2}{x^3+y^3}\le1\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow A\le1\)

Dấu "=" xảy ra khi x=y=1.

Vậy MaxA là 1, đạt được khi x=y=1.

\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5}{16}\left(2x+y\right)\ge2\sqrt{\dfrac{3}{16}.3}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\).

Đẳng thức xảy ra khi x = 1; y = 2.

\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(M=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5\left(2x+y\right)}{16}\ge2\sqrt{\dfrac{9\left(2x+y\right)}{16\left(2x+y\right)}}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{11}{4}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;2\right)\)

Ta có: \(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(=\left(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\right)+\dfrac{5}{8}\dfrac{2x+y}{2}\)

Có: \(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\ge2\sqrt{\dfrac{3}{8}\dfrac{2x+y}{2}\dfrac{3}{2x+y}}=\dfrac{3}{2}\)

Dấu '=' xảy ra <=> \(\dfrac{3}{8}\dfrac{2x+y}{2}=\dfrac{3}{2x+y}\)

Có: \(\dfrac{5}{8}\dfrac{2x+y}{2}\ge\dfrac{5}{8}\sqrt{2xy}=\dfrac{5}{4}\)

Dấu '=' xảy ra <=> 2x=y và xy=2

\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\)

Dấu '=' xảy ra <=> x=1, y=2

Vậy GTNN của M là 11/4 <=> x=1;y=2

Ta có:

\(M=\dfrac{2x+y}{xx}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(=\left(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\right)+\dfrac{5}{8}\dfrac{2x+y}{2}\)

Có: \(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\ge2\sqrt{\dfrac{3}{8}\dfrac{2x+y}{2}\dfrac{3}{2x+y}}=\dfrac{3}{2}\)

Dấu '=' xảy ra \(\Leftrightarrow\dfrac{3}{8}\dfrac{2x+y}{2}=\dfrac{3}{2x+y}\)

Có: \(\dfrac{5}{8}\dfrac{2x+y}{2}\ge\dfrac{5}{8}\sqrt{2xy}=\dfrac{5}{4}\)

Dấu '=' xảy ra \(\Leftrightarrow2x=y,xy=2\)

\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\)

Dấu '=' xảy ra \(\Leftrightarrow x=1,y=2\)

Vậy GTNN của M là \(\dfrac{11}{4}\Leftrightarrow x=1,y=2\)

Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)

\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)

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

Tương tự ta được

\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)

\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :

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

\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)

\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)