Ta cần chứng minh:\(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\sqrt{3}\)

Áp dụng bất đẳng thức Bunhiacopxki, ta được:

\(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\dfrac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\)

Mặt khác, ta có:

\(\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(\left(x+y+xy\right)+\left(y+z+yz\right)+\left(z+x+zx\right)\right)\)

\(\Leftrightarrow\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(6+xy+yz+zx\right)\)Lại có:

\(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{9}{3}=3\)

\(\Rightarrow\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(6+3\right)=27\)

\(\Rightarrow\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\le3\sqrt{3}\)

\(\Rightarrow\dfrac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\ge\dfrac{9}{3\sqrt{3}}=\sqrt{3}\)

Do đó \(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\sqrt{3}\)

Dấu bằng xảy ra \(\Leftrightarrow x=y=z=1\).

trùi s ghim lên đay cx k ai giải v trùi

CHÚC BẠN HỌC TỐT

Thanks

Ta có: \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac52\)

=>\(\frac{x}{\sqrt{xy}}+\frac{y}{\sqrt{xy}}=\frac52\)

=>\(2\left(x+y\right)=5\sqrt{xy}\)

=>\(2x-5\sqrt{xy}+2y=0\)

=>\(2x-4\sqrt{xy}-\sqrt{xy}+2y=0\)

=>\(2\sqrt{x}\left(\sqrt{x}-2\sqrt{y}\right)-\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)=0\)

=>\(\left(\sqrt{x}-2\sqrt{y}\right)\left(2\sqrt{x}-\sqrt{y}\right)=0\)

TH1: \(\sqrt{x}-2\sqrt{y}=0\)

=>\(\sqrt{x}=2\sqrt{y}\)

=>x=4y

\(A=\frac{2x+3\cdot\sqrt{xy}}{2x-3\sqrt{xy}}\)

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

\(=\frac{2\cdot\sqrt{4y}+3\sqrt{y}}{2\cdot\sqrt{4y}-3\sqrt{y}}=\frac{4\sqrt{y}+3\sqrt{y}}{4\sqrt{y}-3\sqrt{y}}=\frac{4+3}{4-3}=7\)

TH2: \(2\sqrt{x}-\sqrt{y}=0\)

=>\(\sqrt{y}=2\sqrt{x}\)

=>y=4x

\(A=\frac{2x+3\cdot\sqrt{xy}}{2x-3\sqrt{xy}}\)

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

\(=\frac{2\sqrt{x}+3\cdot\sqrt{4x}}{2\sqrt{x}-3\cdot\sqrt{4x}}=\frac{2\sqrt{x}+6\sqrt{x}}{2\sqrt{x}-6\sqrt{x}}=\frac{8}{-4}=-2\)

Bài 2 xét x=0 => A =0

xét x>0 thì \(A=\frac{1}{x-2+\frac{2}{\sqrt{x}}}\)

để A nguyên thì \(x-2+\frac{2}{\sqrt{x}}\inƯ\left(1\right)\)

=>cho \(x-2+\frac{2}{\sqrt{x}}\)bằng 1 và -1 rồi giải ra =>x=?

1,Ta có \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\)

=> \(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)

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

\(b+2=\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)\)

\(c+2=\left(\sqrt{c}+\sqrt{b}\right)\left(\sqrt{c}+\sqrt{a}\right)\)

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

=> \(\frac{\sqrt{a}}{a+2}+...=\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\frac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)

=> M=0

Vậy M=0 

\(3=x+y+xy\le\sqrt{2\left(x^2+y^2\right)}+\dfrac{x^2+y^2}{2}\)

\(\Rightarrow\left(\sqrt{x^2+y^2}-\sqrt{2}\right)\left(\sqrt{x^2+y^2}+3\sqrt{2}\right)\ge0\)

\(\Rightarrow x^2+y^2\ge2\)

\(\Rightarrow-\left(x^2+y^2\right)\le-2\)

\(P=\sqrt{9-x^2}+\sqrt{9-y^2}+\dfrac{x+y}{4}\le\sqrt{2\left(9-x^2+9-y^2\right)}+\dfrac{\sqrt{2\left(x^2+y^2\right)}}{4}\)

\(P\le\sqrt{2\left(18-x^2-y^2\right)}+\dfrac{1}{4}.\sqrt{2\left(x^2+y^2\right)}\)

\(P\le\left(\sqrt{2}-1\right)\sqrt{18-x^2-y^2}+\sqrt[]{2}\sqrt{\dfrac{\left(18-x^2-y^2\right)}{2}}+\dfrac{1}{2}\sqrt{\dfrac{x^2+y^2}{2}}\)

\(P\le\left(\sqrt{2}-1\right).\sqrt{18-2}+\sqrt{\left(2+\dfrac{1}{4}\right)\left(\dfrac{18-x^2-y^2+x^2+y^2}{2}\right)}=\dfrac{1+8\sqrt{2}}{2}\)

Dấu "=" xảy ra khi \(x=y=1\)