Ta có: A = \(x+\frac{9}{x-1}+3=x-1+\frac{9}{x-1}+4\)

Áp dụng bđt cosi cho các số dương x - 1 và 9/(x - 1) Ta có: (x > 3)

\(x-1+\frac{9}{x-1}\ge2.\sqrt{\left(x-1\right)\cdot\frac{9}{x-1}}=2.3=6\)

=> A \(\ge\)6 + 4 = 10

Dấu "=" xảy ra <=> \(x-1=\frac{9}{x-1}\) <=> (x - 1)² = 9 <=> \(\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}\) <=> \(\orbr{\begin{cases}x=4\left(tm\right)\\x=-2\left(ktm\right)\end{cases}}\)

Vậy MinA = 10 <=> x = 4

ĐKXĐ x > 1

\(M=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

\(M=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)

\(M=\frac{\sqrt{x}-1}{\sqrt{x}}\)

ĐKXĐ x>1

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

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

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

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

Áp dụng bất đẳng thức Cauchy-Schwarz ta có:

\(\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{a}.\sqrt{a}+\sqrt{b}.\sqrt{c}\)

\(\Leftrightarrow\sqrt{\left(a+b\right)\left(a+c\right)}\ge a+\sqrt{bc}\)

Do đó \(\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}=\frac{\sqrt{bc\left(c+a\right)\left(a+b\right)}}{\left(c+a\right)\left(a+b\right)}\ge\sqrt{abc}\frac{\sqrt{a}}{\left(c+a\right)\left(c+b\right)}+\frac{bc}{\left(c+a\right)\left(c+b\right)}\left(1\right)\)

Chứng minh tương tự ta được: 

\(\hept{\begin{cases}\sqrt{\frac{bc}{\left(c+b\right)\left(a+b\right)}}=\frac{\sqrt{bc\left(c+b\right)\left(a+b\right)}}{\left(c+b\right)\left(a+b\right)}\ge\sqrt{abc}\frac{\sqrt{b}}{\left(c+b\right)\left(a+b\right)}+\frac{ac}{\left(c+b\right)\left(a+b\right)}\left(2\right)\\\sqrt{\frac{ca}{\left(c+a\right)\left(a+b\right)}}=\frac{\sqrt{ca\left(c+a\right)\left(a+b\right)}}{\left(c+a\right)\left(a+b\right)}\ge\sqrt{abc}\frac{\sqrt{c}}{\left(c+a\right)\left(a+b\right)}+\frac{ab}{\left(a+c\right)\left(a+b\right)}\left(3\right)\end{cases}}\)

\(\Rightarrow\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(c+b\right)\left(a+b\right)}}+\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\ge\)

\(\sqrt{abc}\left(\frac{\sqrt{a}}{\left(a+c\right)\left(a+b\right)}+\frac{\sqrt{b}}{\left(c+b\right)\left(a+b\right)}+\frac{\sqrt{c}}{\left(c+b\right)\left(a+c\right)}\right)+\)\(\frac{bc}{\left(a+c\right)\left(a+b\right)}+\frac{ac}{\left(c+b\right)\left(a+b\right)}+\frac{ab}{\left(c+b\right)\left(a+c\right)}\left(4\right)\)

Ta lại có: \(\frac{bc}{\left(a+c\right)\left(a+b\right)}+\frac{ac}{\left(c+b\right)\left(a+b\right)}+\frac{ab}{\left(c+b\right)\left(a+c\right)}+\frac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(=\frac{bc\left(b+c\right)+ac\left(a+c\right)+ab\left(a+b\right)+2abc}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}\)

\(=\frac{bc\left(a+b+c\right)+ca\left(a+b+c\right)+ab\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{c\left(a+b+c\right)\left(b+a\right)+ab\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(=\frac{\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{\left(a+b\right)\left(c+b\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1\)

\(\left(4\right)\Leftrightarrow\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(c+b\right)\left(a+b\right)}}+\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\)\(\ge\sqrt{abc}\left(\frac{\sqrt{a}}{\left(c+a\right)\left(a+b\right)}+\frac{\sqrt{b}}{\left(c+b\right)\left(a+b\right)}+\frac{\sqrt{c}}{\left(c+b\right)\left(a+c\right)}\right)+1-\frac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Do đó ta cần chứng minh \(\sqrt{abc}\left(\frac{\sqrt{a}}{\left(c+a\right)\left(a+b\right)}+\frac{\sqrt{b}}{\left(c+b\right)\left(a+b\right)}+\frac{\sqrt{c}}{\left(c+b\right)\left(a+c\right)}\right)+1-\frac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)\(\ge1+\frac{4abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Điều này tương đương với \(\sqrt{a}\left(b+c\right)+\sqrt{b}\left(a+c\right)+\sqrt{c}\left(a+b\right)\ge6\sqrt{abc}\left(5\right)\)

Theo bất đẳng thức AM-GM thì (5) luôn đúng

Dấu "=" xảy ra khi (1);(2);(3) và (5) xảy ra dấu "=". điều này tương đương với a=b=c

Vậy ta có điều phải chứng minh

=))