Lời giải:
\(S=a^2+\frac{18}{\sqrt{a}}=a^2(1-\frac{1}{2\sqrt{6}})+\frac{a^2}{2\sqrt{6}}+\frac{18}{\sqrt{a}}\)

Áp dụng BĐT AM-GM:

\(\frac{a^2}{2\sqrt{6}}+\frac{18}{\sqrt{a}}\geq 2\sqrt{\frac{3\sqrt{6}.\sqrt{a^3}}{2}}\geq 2\sqrt{\frac{3\sqrt{6}.\sqrt{6^3}}{2}}=6\sqrt{6}\) (do $a\geq 6$)

\(a^2(1-\frac{1}{2\sqrt{6}}\geq 6^2(1-\frac{1}{2\sqrt{6}})=36-3\sqrt{6}\) (do $a\geq 6$)

Cộng lại:

\(\Rightarrow S\ge 36+3\sqrt{6}\)

Vậy $S_{\min}=36+3\sqrt{6}$ khi $a=6$

Xét \(\left(a^2+\frac{1}{b+c}\right)\left(4^2+1^2\right)\ge\left(4a+\frac{1}{\sqrt{b+c}}\right)^2\)

=> \(\sqrt{a^2+\frac{1}{b+c}}\ge\frac{4a+\frac{1}{\sqrt{b+c}}}{\sqrt{17}}\)

Tương tự => \(\left\{{}\begin{matrix}\sqrt{b^2+\frac{1}{c+a}}\ge\frac{4b+\frac{1}{\sqrt{c+a}}}{\sqrt{17}}\\\sqrt{c^2+\frac{1}{a+b}}\ge\frac{4c+\frac{1}{\sqrt{a+b}}}{\sqrt{17}}\end{matrix}\right.\)

=> A \(\ge\frac{4\left(a+b+c\right)+\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}}{\sqrt{17}}\)

Có \(\frac{1}{\sqrt{a+b}}=\frac{4}{4.\sqrt{a+b}}\)

Mà \(\sqrt{\left(a+b\right).4}\le\frac{a+b+4}{2}\) => \(4\sqrt{a+b}\le a+b+4\)

=> \(\frac{1}{\sqrt{a+b}}\ge\frac{4}{a+b+4}\)

Tương tự => \(\left\{{}\begin{matrix}\frac{1}{\sqrt{b+c}}\ge\frac{4}{b+c+4}\\\frac{1}{\sqrt{c+a}}\ge\frac{4}{c+a+4}\end{matrix}\right.\)

=> \(\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}\) \(\ge4.\left(\frac{1}{b+c+4}+\frac{1}{c+a+4}+\frac{1}{a+b+4}\right)\)

\(\ge4.\frac{9}{2a+2b+2c+12}=\frac{3}{2}\)

=> \(A\ge\frac{4.6+\frac{3}{2}}{\sqrt{17}}=\frac{3.\sqrt{17}}{2}\)

a) Điều kiện xác định : \(a>0\); \(a\ne1\)

b) Ta có : 

\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}=\left(\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}.\left(\sqrt{a}+1\right)}\right).\frac{a}{\sqrt{a}+1}\)

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

\(=\sqrt{a}.\left(\sqrt{a}-1\right)=a-\sqrt{a}\)

c)

Ta có :  \(A=a-\sqrt{a}=\left(a-2.\frac{1}{2}.\sqrt{a}+\frac{1}{4}\right)-\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\)

Vì \(a>0\)và  \(a\ne1\)nên \(\left(\sqrt{a}-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\)  \(A=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Vậy \(Min_A=-\frac{1}{4}\) khi và chỉ khi \(\sqrt{a}-\frac{1}{2}=0\Rightarrow\sqrt{a}=\frac{1}{2}\Rightarrow a=\frac{1}{4}\)

\(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

\(S=\frac{a^2+b^2}{ab}\ge\frac{2ab}{ab}=2\)

Dấu \(=\)xảy ra khi \(a=b\).

Vậy \(minS=2\).

\(S=\frac{a^2+b^2}{ab}=\frac{a^2}{ab}+\frac{b^2}{ab}\ge\frac{\left(a+b\right)^2}{2ab}\)( Cauchy-Schwarz dạng Engel )

Lại có : \(2ab\le\frac{\left(a+b\right)^2}{2}\)( AM-GM )

\(\Rightarrow\frac{1}{2ab}\ge\frac{1}{\frac{\left(a+b\right)^2}{2}}=\frac{2}{\left(a+b\right)^2}\Rightarrow\frac{\left(a+b\right)^2}{2ab}\ge2\)

Dấu "=" xảy ra <=> a = b

Vậy MinS = 2

xin lỗi nha nhập thiếu đề 

a>= 2b 

Chứ nếu nó kiểu kia thì khỏi có a>= b làm gì???

ĐỀ THI HỌC SINH GIỎI MÀ 

dùng đạo hàm thì đc min =5/2 thì phải 

Ta có : \(\left\{{}\begin{matrix}x\ge1\\y\ge2\\z\ge3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}\ge0\\\sqrt{y-2}\ge0\\\sqrt{z-3}\ge0\end{matrix}\right.\Rightarrow\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}\ge0\)

Đặt \(\sqrt{x-1}=a;\sqrt{y-2}=b;\sqrt{z-3}=c\)

\(\Rightarrow A=\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\)

\(\sum\frac{a}{a^2+1}=\sum\left(a-\frac{a^3}{a^2+1}\right)\ge\sum\left(a-\frac{a}{2}\right)=\frac{a+b+c}{2}\)

\(\Rightarrow A\ge\frac{\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}}{2}=0\)

Vậy \(MIN_A=0\) khi \(x=1;y=2;z=3\)

\(A=\frac{1.\sqrt{x-1}}{x}+\frac{1}{\sqrt{2}}.\frac{\sqrt{2}.\sqrt{y-2}}{y}+\frac{1}{\sqrt{3}}.\frac{\sqrt{3}.\sqrt{z-3}}{z}\)

\(A\ge\frac{1+x-1}{2x}+\frac{1}{\sqrt{2}}\left(\frac{2+y-2}{2y}\right)+\frac{1}{\sqrt{3}}\left(\frac{3+z-3}{2z}\right)=\frac{6+3\sqrt{2}+2\sqrt{3}}{12}\)

\(\Rightarrow A_{min}=\frac{6+3\sqrt{2}+2\sqrt{3}}{12}\) khi \(\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-2}=\sqrt{2}\\\sqrt{z-3}=\sqrt{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)

A=
x
1.
x−1
​

​
+
2
​

1
​
.
y
2
​
.
y−2
​

​
+
3
​

1
​
.
z
3
​
.
z−3
​

​


A\ge\frac{1+x-1}{2x}+\frac{1}{\sqrt{2}}\left(\frac{2+y-2}{2y}\right)+\frac{1}{\sqrt{3}}\left(\frac{3+z-3}{2z}\right)=\frac{6+3\sqrt{2}+2\sqrt{3}}{12}A≥
2x
1+x−1
​
+
2
​

1
​
(
2y
2+y−2
​
)+
3
​

1
​
(
2z
3+z−3
​
)=
12
6+3
2
​
+2
3
​

​


\Rightarrow A_{min}=\frac{6+3\sqrt{2}+2\sqrt{3}}{12}⇒A
min
​
=
12
6+3
2
​
+2
3
​

​
khi \left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-2}=\sqrt{2}\\\sqrt{z-3}=\sqrt{3}\end{matrix}\right.
⎩
⎨
⎧
​

x−1
​
=1
y−2
​
=
2
​

z−3
​
=
3
​

​
\Rightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.⇒
⎩
⎨
⎧
​

x=2
y=4
z=6
​

tìm GTLN