\(A\le\sqrt{3\left(x+y+y+z+z+x\right)}=\sqrt{6\left(x+y+z\right)}\le\sqrt{6.\sqrt{3\left(x^2+y^2+z^2\right)}}=\sqrt{6\sqrt{3}}\)

\(A_{max}=\sqrt{6\sqrt{3}}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)

Do \(x^2+y^2+z^2=1\Rightarrow0\le x;y;z\le1\)

\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\\z^2\le z\end{matrix}\right.\) \(\Rightarrow x+y+z\ge x^2+y^2+z^2=1\)

\(A^2=2\left(x+y+z\right)+2\sqrt{\left(x+y\right)\left(x+z\right)}+2\sqrt{\left(x+y\right)\left(y+z\right)}+2\sqrt{\left(y+z\right)\left(z+x\right)}\)

\(A^2=2\left(x+y+z\right)+2\sqrt{x^2+xy+yz+zx}+2\sqrt{y^2+xy+yz+zx}+2\sqrt{z^2+xy+yz+zx}\)

\(A^2\ge2\left(x+y+z\right)+2\sqrt{x^2}+2\sqrt{y^2}+2\sqrt{z^2}=4\left(x+y+z\right)\ge4\)

\(\Rightarrow A\ge2\)

\(A_{min}=2\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị

Ta có:(x+1)²>0

=>(x+1)²+4>4

=>GTNN (x+1)²+4 là 4 <=>(x+1)²=<=>x+1=0<=>x=-1

\(\left(x+1\right)^2+4\)

mà  \(\left(x+1\right)^2\ge0\)

nên GTNN của \(\left(x+1\right)^2+4\)\(=4\)  tại  \(x=-1\)

(x+1)²+4

mà (x+1)² \(\ge0\)

=> biểu thức đạt GTNN là 4

<=> x+1=0

=> x=0-1

=> x=-1

ta thấy 1+x>= 2 căn x

=> 2 căn x/1+x bé hơn hoặc = 1

hok tốt

dấu = xảy ra khi x=-1

ĐKXĐ: x > 0

Áp dụng bđt Cô-si có \(x+1\ge2\sqrt{x}\)

                              \(\Rightarrow\frac{2\sqrt{x}}{1+x}\le1\)

Dấu "=" tại x = 1 (T/m ĐKXĐ)

\(\)

\(p=\sqrt{x}\left(1-\sqrt{x}\right)=-x+\sqrt{x}=-\left(\sqrt{x}\right)^2+2\sqrt{x}\cdot\frac{1}{2}-\frac{1}{4}+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\Leftrightarrow x=\sqrt{\frac{1}{2}}\)

Đặt:

\(A=2x^2-6x\)

\(A=2x^2-6x+\dfrac{9}{2}-\dfrac{9}{2}\)

\(A=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}\)

\(A=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)

Vì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\) nên \(A=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Dấu "=" xảy ra khi:

\(x=-\dfrac{3}{2}\)

\(2x^2-6x\)

\(=2.\left(x^2-3x\right)\)

=\(2\left[x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3^{ }}{2}\right)^2-\left(\dfrac{3}{2}\right)^2\right]\)

\(=2\left[\left(x-\dfrac{3}{2}\right)^2-\left(\dfrac{3}{2}\right)^2\right]\)

=\(2\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]\ge2\left(0-\dfrac{9}{4}\right)\ge0\)

Vậy GTNN của biểu thức là\(\dfrac{-9}{2}\) xẩy ra khi \(x=\dfrac{3}{2}\)

Nguồn: OLM

Bạn học tốt nhé!

\(2x^2-6x\\ =2x^2-6x+\dfrac{9}{2}-\dfrac{9}{2}\\ =2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ =2\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\dfrac{9}{2}\\ =2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)

\(\text{Ta có: }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\forall x\)

Dấu \("="\) xảy ra khi:

\(2\left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow \left(x-\dfrac{3}{2}\right)^2=0\\\Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\)

Vậy \(GTNN\) của biểu thức là \(-\dfrac{9}{2}\) khi \(x=\dfrac{3}{2}\)