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 \(x=y=\dfrac{1}{2}\)

Áp dụng BĐT BSC:

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

\(\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}\right)\)

\(=\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{4}.4=1\)

\(maxF=1\Leftrightarrow x=y=z=\dfrac{3}{4}\)

\(P=\dfrac{x+2y}{2xy}+\dfrac{1}{x+2y}=\dfrac{x+2y}{4}+\dfrac{1}{x+2y}\)

\(P=\dfrac{x+2y}{16}+\dfrac{1}{x+2y}+\dfrac{3\left(x+2y\right)}{16}\)

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

\(P_{min}=\dfrac{5}{4}\) khi \(\left(x;y\right)=\left(2;1\right)\)

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

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 \(\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\Rightarrow\dfrac{y}{x}\ge4\)

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

Đặt \(\dfrac{y}{x}=a\ge4\Rightarrow P=\dfrac{2a^2-2a+1}{a+1}=2a-4+\dfrac{5}{a+1}\)

\(P=\dfrac{a+1}{5}+\dfrac{5}{a+1}+\dfrac{9}{5}.a-\dfrac{21}{5}\ge2\sqrt{\dfrac{5\left(a+1\right)}{5\left(a+1\right)}}+\dfrac{9}{5}.4-\dfrac{21}{5}=5\)

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

đây là những món quà mà bn sẽ nhận đc: 1: áo quần 2: tiền 3: đc nhiều người yêu quý 4: may mắn cả 5: luôn vui vẻ trong cuộc sống 6: đc crush thích thầm 7: học giỏi 8: trở nên xinh đẹp phật sẽ ban cho bn những điều này nếu cậu gửi tin nhắn này cho 25 người,

\(x+2y=6\)

\(\Leftrightarrow\dfrac{6}{2}=\dfrac{x}{2}+y\)

\(P+\dfrac{6}{2}=\dfrac{8}{x}+\dfrac{1}{y}+\dfrac{x}{2}+y\)

\(\Leftrightarrow P+\dfrac{6}{2}=\left(\dfrac{8}{x}+\dfrac{1}{y}\right)+\left(\dfrac{1}{y}+y\right)\)

vì x;y là số thực dương ,áp dụng BĐT Côsi ta có :

\(\dfrac{8}{x}+\dfrac{x}{2}=2\sqrt{\dfrac{8}{x}+\dfrac{x}{2}}=2\sqrt{4}=2.2=4\)

\(\dfrac{1}{y}+y=2\sqrt{\dfrac{1}{y}+y}=2\sqrt{1}=2.1=2\)

nên \(P+\dfrac{6}{2}\ge6\)

\(\Leftrightarrow P\ge6-\dfrac{6}{2}\)

\(\Leftrightarrow P\ge3\)

vậy \(P_{min}=3\)