đề là GTLN.

ĐKXĐ : \(3\le x\le5\)

Ta có : \(A^2=\left(\sqrt{x-3}+\sqrt{5-x}\right)^2=x-3+5-x+2\sqrt{\left(x-3\right)\left(5-x\right)}\)

\(A^2=2+2\sqrt{\left(x-3\right)\left(5-x\right)}\le2+\left(x-3+5-x\right)=4\)

\(\Rightarrow\)A² max = 4 \(\Rightarrow\)A max = 2 \(\Leftrightarrow\) x = 4

ĐKXĐ: \(3\le x\le5\)

Dễ thấy \(A\ge0\). Xét : \(A^2=\left(\sqrt{x-3}+\sqrt{5-x}\right)^2\)

                                            \(=x-3+2\sqrt{\left(x-3\right)\left(5-x\right)}+5-x\)

                                           \(=2+2\sqrt{\left(x-3\right)\left(5-x\right)}\)

Vì \(\sqrt{\left(x-3\right)\left(5-x\right)}\ge0\Rightarrow2\sqrt{\left(x-3\right)\left(5-x\right)}\ge0\)

Hay \(A^2\ge2+0=2\Rightarrow A\ge\sqrt{2}.\)

Vậy Giá trị nhỏ nhất của biểu thức \(A=\sqrt{2}\)Khi \(\sqrt{\left(x-3\right)\left(5-x\right)}=0\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)

với a,b,c lớn thì \(\frac{1}{\left(a+2b+c\right)^3}\) nhỏ, \(a^3+8b^3+c^3\) lớn => P ko có max 

\(P=\frac{a^3+8b^3+c^3}{\left(a+2b+c\right)^3}=\left(\frac{a}{a+2b+c}\right)^3+\left(\frac{2b}{a+2b+c}\right)^3+\left(\frac{c}{a+2b+c}\right)^3\)

Đặt \(\left(x;y;z\right)\rightarrow\left(\frac{a}{a+2b+c};\frac{2b}{a+2b+c};\frac{c}{a+2b+c}\right)\)\(\Rightarrow\)\(x+y+z=1\)

\(P=x^3+y^3+z^3=\frac{x^4}{x}+\frac{y^4}{y}+\frac{z^4}{z}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\frac{\frac{\left(x+y+z\right)^4}{9}}{x+y+z}=\frac{1}{9}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{3}\) hay \(a=2b=c\)

\(A\ge3\left(a+b+c\right)+\frac{9}{a+b+c}=3.3+\frac{9}{3}=12\)

\(A_{min}=12\) khi \(a=b=c=1\)

 Ta cần chứng minh: \(3a+\frac{1}{a}\ge2a+2\Leftrightarrow3a+\frac{1}{a}-4\ge2\left(a-1\right)\)

\(\Leftrightarrow\frac{3a^2-4a+1}{a}-2\left(a-1\right)\ge0\Leftrightarrow\left(a-1\right)\left(\frac{3a-1}{a}-2\right)\ge0\Leftrightarrow\frac{\left(a-1\right)^2}{a}\)(đúng)

Tương tự: \(3b+\frac{1}{b}\ge2b+2;3c+\frac{1}{c}\ge2c+2\)

Cộng theo vế: \(A\ge2\left(a+b+c\right)+6=12\)

Dấu bằng xảy ra khi a=b=c=1