Ta có \(\frac{a^2\left(b+1\right)}{a+b+ab}=\frac{a\left(ab+a+b\right)-ab}{ab+a+b}=a-\frac{ab}{ab+a+b}\)

Mà \(\frac{1}{ab+b+a}\le\frac{1}{9}\left(\frac{1}{ab}+\frac{1}{b}+\frac{1}{a}\right)\)

=> \(\frac{a^2\left(b+1\right)}{a+b+ab}\ge a-\frac{1}{9}ab\left(\frac{1}{ab}+\frac{1}{a}+\frac{1}{b}\right)=\frac{8}{9}a-\frac{1}{9}b-\frac{1}{9}\)

=> \(VT\ge\frac{7}{9}\left(a+b+c\right)-\frac{1}{3}=\frac{7}{3}-\frac{1}{3}=2\)

MinVT=2  khi a=b=c=1

Áp dụng: (a + b)² ≥ 4ab Ta có: 
(x + y + z)² ≥ 4(x + y)z hay 1 ≥ 4(x + y)z (*)        (Vì x + y + z = 1) 
=> (x + y)/xyz ≥ 4(x + y)²z/xyz      ( Nhân hai vế (*) với (x + y)/xyz) 
=> (x + y)/xyz ≥ 4.4xyz/xyz = 16    (vì (x + y)² ≥ 4xy) 
Vậy min A = 16 <=> x = y; x + y = z và x + y + z = 1 
=> x = y = 1/4; z = 1/2

bn Phùng Gia Bảo nhầm 1 chỗ r nhe

C1: \(A=\frac{x+y+z}{xyz}=\frac{1}{\left(\sqrt[3]{xyz}\right)^3}\ge\frac{1}{\left(\frac{x+y+z}{3}\right)^3}=\frac{1}{\frac{1}{27}}=27\)

C2: \(A=\frac{x+y+z}{xyz}=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ge\frac{9}{xy+yz+zx}\ge\frac{9}{\frac{\left(x+y+z\right)^2}{3}}=\frac{9}{\frac{1}{3}}=27\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{3}\)

\(1,A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)

                                             \(\ge\frac{4}{\left(x+y^2\right)}+\frac{1}{\frac{\left(x+y\right)^2}{2}}\ge\frac{4}{1}+\frac{2}{1}=6\)

Dấu "=" <=> x= y = 1/2

\(2,A=\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\left(\frac{x}{9y}+\frac{y}{x}\right)+\frac{8x}{9y}\ge2\sqrt{\frac{x}{9y}.\frac{y}{x}}+\frac{8.3y}{9y}\)

                                                                                                  \(=2\sqrt{\frac{1}{9}}+\frac{8.3}{9}=\frac{10}{3}\)

Dấu "=" <=> x = 3y

\(A=\frac{7}{3+\sqrt{2}}+\frac{2}{1-\sqrt{3}}\)

\(=\frac{7\left(3-\sqrt{2}\right)}{3^2-\sqrt{2}^2}+\frac{2\left(1+\sqrt{3}\right)}{1^2-\sqrt{3}^2}\)

\(=\frac{7\left(3-\sqrt{2}\right)}{7}+\frac{2\left(1+\sqrt{3}\right)}{-2}\)

\(=3-\sqrt{2}-1-\sqrt{3}\)

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

\(A=\frac{7}{3+\sqrt{2}}+\frac{2}{1-\sqrt{3}}=\frac{7\left(3-\sqrt{2}\right)}{3^2-\left(\sqrt{2}\right)^2}+\frac{2\left(1+\sqrt{3}\right)}{1^2-\left(\sqrt{3}\right)^2}\)

\(=\frac{7\left(3-\sqrt{2}\right)}{9-2}+\frac{2\left(1+\sqrt{3}\right)}{1-3}=\frac{7\left(3-\sqrt{2}\right)}{7}+\frac{2\left(1+\sqrt{3}\right)}{-2}\)

\(=\left(3-\sqrt{2}\right)-\left(1+\sqrt{3}\right)=3-\sqrt{2}-1-\sqrt{3}=2-\sqrt{2}-\sqrt{3}\)

\(\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{6}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{\left(\sqrt{2}+\sqrt{3}\right)^2-\sqrt{6}^2}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{5+2\sqrt{6}-6}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{2\sqrt{6}-1}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{6}+1\right)}{2\sqrt{6}^2-1^2}\)

\(=\frac{4\sqrt{3}+6\sqrt{2}+12+\sqrt{2}+\sqrt{3}+\sqrt{6}}{11}\)

\(=\frac{\sqrt{6}+5\sqrt{3}+7\sqrt{2}+12}{11}\)

\(\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2\) \(ĐKXĐ:\hept{\begin{cases}a\ge0\\b\ge0\\a\ne b\end{cases}}\)

\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)

\(=\left(\left(a+\sqrt{ab}+b\right)+\sqrt{ab}\right)\left(\frac{1}{\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)

\(=1\)

\(\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)

\(=\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)

\(=\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)

\(=\left(\frac{a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)

\(=\left(\frac{a\left(\sqrt{a}+\sqrt{b}\right)-b\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)

\(=\left(\frac{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)

\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)

\(=\left(\sqrt{a}+\sqrt{b}\right)^2\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)

\(=\left(\sqrt{a}+\sqrt{b}\right)^2\left(\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2.\)

\(=\left(\sqrt{a}+\sqrt{b}\right)^2\cdot\frac{1}{\left(\sqrt{a}+\sqrt{b}\right)^2}.\)\(=1\)

\(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\) 

= \(\frac{\sqrt{3}.\left(\sqrt{1+\sqrt{3}}+1\right)}{\left(\sqrt{\sqrt{3}+1}-1\right)\left(\sqrt{\sqrt{3}+1}+1\right)}-\frac{\sqrt{3}.\left(\sqrt{1+\sqrt{3}}-1\right)}{\left(\sqrt{\sqrt{3}+1}-1\right)\left(\sqrt{\sqrt{3}+1}-1\right)}\)

= \(\frac{\sqrt{3}.\left(\sqrt{1+\sqrt{3}}+1\right)-\sqrt{3}.\left(\sqrt{1+\sqrt{3}}-1\right)}{\left(\sqrt{1+\sqrt{3}}+1\right)\left(\sqrt{1+\sqrt{3}}-1\right)}\)

= \(\frac{\sqrt{3}.\left(\sqrt{1+\sqrt{3}}+1\right)-\sqrt{3}.\left(\sqrt{1+\sqrt{3}}-1\right)}{\sqrt{3}}\)

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

= 2