\(pt< =>\hept{\begin{cases}x+y+2\sqrt{xy}=4\\x+y+6+2\sqrt{\left(x+3\right)\left(y+3\right)}=16\end{cases}}\)

<=>\(\hept{\begin{cases}x+y=4-2\sqrt{xy}\\x+y=10-2\sqrt{\left(x+3\right)\left(y+3\right)}\end{cases}}\)

=> \(4-2\sqrt{xy}=10-2\sqrt{\left(x+3\right)\left(y+3\right)}\)

<=>\(-2\sqrt{xy}=6-2\sqrt{\left(x+3\right)\left(y+3\right)}\)

<=> \(\sqrt{\left(x+3\right)\left(y+3\right)}=\sqrt{xy}+3\)

Bình phương hai vế, tự làm nốt

Lấy tổng, tích ta được:

\(\hept{\begin{cases}\sqrt{x+3}-\sqrt{x}+\sqrt{y+3}-\sqrt{y}=2\\\sqrt{x+3}+\sqrt{y}+\sqrt{y+3}+\sqrt{y}=6\end{cases}}\)Đặt \(\hept{\begin{cases}\sqrt{x+3}+\sqrt{x}=a\left(a>0\right)\\\sqrt{y+3}+\sqrt{y}=b\left(b>0\right)\end{cases}}\)và chú ý rằng \(\hept{\begin{cases}\sqrt{x+3}-\sqrt{x}=\frac{3}{a}\\\sqrt{y+3}-\sqrt{y}=\frac{3}{b}\end{cases}}\)

=>\(\hept{\begin{cases}a+b=6\\\frac{3}{a}+\frac{3}{b}=2\ge\frac{3.4}{a+b}=2\end{cases}}\)(theo Cauchy scharws)

Dấu bằng khi a=b=3

<=>x=y=1

nhan ca  tu va mau voi\(\sqrt{2}\) ta dc

\(\frac{\sqrt{2x-4\sqrt{2x-4}}}{2}=\frac{\sqrt{2x-4-4\sqrt{2x-4}}}{2}=\frac{\sqrt{\left(\sqrt{2x-4}-2\right)^2}}{2}\)(dkx>=2)

                                                   =\(\frac{\left|\sqrt{2x-4}-2\right|}{2}\)

có viết đb đúng ko thế

ĐK  ; \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

a, \(Q=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x-8\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-7\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-7}{\sqrt{x}+1}\)

b. \(Q< \frac{1}{2}\Rightarrow\frac{\sqrt{x}-7}{\sqrt{x}+1}-\frac{1}{2}< 0\Rightarrow\frac{\sqrt{x}-15}{2\left(\sqrt{x}+1\right)}< 0\Rightarrow\sqrt{x}-15< 0\)

\(\Rightarrow0\le x< 225\)và \(x\ne4\)

c. \(Q=\frac{\sqrt{x}-7}{\sqrt{x}+1}=1-\frac{8}{\sqrt{x}+1}\)

Ta thấy \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\frac{-8}{\sqrt{x}+1}\ge-8\Rightarrow1-\frac{8}{\sqrt{x}+1}\ge-7\)

\(\Rightarrow Q\ge-7\)

Vậy \(MinQ=-7\). Dấu bằng xảy ra \(\Rightarrow x=0\)

tương tự Xem câu hỏi

JBMO 2003 :3

Dễ thấy:\(y\le\frac{y^2+1}{2}\Rightarrow\frac{1+x^2}{1+y+z^2}\ge\frac{1+x^2}{1+z^2+\frac{1+y^2}{2}}\)

Tương tự cho các BĐT còn lại nhé :v

Đặt \(a=1+x^2;b=1+y^2;c=1+z^2\) thì cần cm

\(A=\frac{a}{2c+b}+\frac{b}{2a+c}+\frac{c}{2b+a}\ge1\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có: 

\(A=\frac{a^2}{2ac+ab}+\frac{b^2}{2ab+bc}+\frac{c^2}{2bc+ac}\)

\(\ge\frac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\frac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\)