Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!

Có \(\sqrt{\dfrac{xy}{x+y+2z}}=\dfrac{\sqrt{xy}}{\sqrt{x+y+2z}}\)\(=\dfrac{2\sqrt{xy}}{\sqrt{\left(1+1+2\right)\left(x+y+2z\right)}}\)\(\le\dfrac{2\sqrt{xy}}{\sqrt{x}+\sqrt{y}+2\sqrt{z}}\) (theo bunhia dưới mẫu)\(\le\dfrac{2\sqrt{xy}}{4}\left(\dfrac{1}{\sqrt{x}+\sqrt{z}}+\dfrac{1}{\sqrt{y}+\sqrt{z}}\right)\)

\(\Leftrightarrow\sqrt{\dfrac{xy}{x+y+2z}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}}{\sqrt{y}+\sqrt{z}}\right)\)

Tương tự cũng có:

\(\sqrt{\dfrac{yz}{y+z+2x}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{yz}}{\sqrt{y}+\sqrt{x}}+\dfrac{\sqrt{yz}}{\sqrt{z}+\sqrt{x}}\right)\)

\(\sqrt{\dfrac{zx}{z+x+2y}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{zx}}{\sqrt{z}+\sqrt{y}}+\dfrac{\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)

Cộng vế với vế ta được:

 \(VT\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}+\sqrt{yz}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}+\sqrt{zx}}{\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{yz}+\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)

\(\Leftrightarrow VT\le\dfrac{1}{2}\left(\sqrt{y}+\sqrt{x}+\sqrt{z}\right)=\dfrac{1}{2}\)

Dấu = xảy ra khi \(x=y=z=\dfrac{1}{9}\)

hay

Lời giải:
Từ đkđb suy ra:
$x-y=\frac{1}{z}-\frac{1}{y}=\frac{y-z}{yz}$

$y-z=\frac{1}{x}-\frac{1}{z}=\frac{z-x}{xz}$

$z-x=\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}$

$\Rightarrow (x-y)(y-z)(z-x)=\frac{(y-z)(z-x)(x-y)}{(xyz)^2}$

$\Leftrightarrow (x-y)(y-z)(z-x)(1-\frac{1}{x^2y^2z^2})=0$

$\Rightarrow (x-y)(y-z)(z-x)=0$ hoặc $1-\frac{1}{x^2y^2z^2}=1$

$\Rightarrow (x-y)(y-z)(z-x)=0$ hoặc $x^2y^2z^2=1$
Nếu $(x-y)(y-z)(z-x)=0$

$\Rightarrow x=y$ hoặc $y=z$ hoặc $z=x$

Không mất tquat giả sử $x=y$. Khi đó: $\frac{1}{y}=\frac{1}{z}$

$\Rightarrow y=z$

$\Rightarrow x=y=z$. Tương tự khi xét $y=z$ hoặc $z=x$ thì ta cũng thu được $x=y=z$
Vậy $x=y=z$ hoặc $x^2y^2z^2=1$

Haha không giỡn nữa :v

Áp dụng BĐT Cauchy-Schwarz ta có:

\(L.H.S=Σ\dfrac{1}{2x+y+z}=7Σ\dfrac{1}{2\left(x+3y\right)+\left(y+3z\right)+4\left(z+3x\right)}\)

\(=\dfrac{1}{7}Σ\dfrac{\left(2+1+4\right)^2}{2\left(x+3y\right)+\left(y+3z\right)+4\left(z+3x\right)}\)

\(\le\dfrac{1}{7}Σ\left(\dfrac{2^2}{2\left(x+3y\right)}+\dfrac{1^2}{y+3z}+\dfrac{4^2}{4\left(z+3x\right)}\right)\)

\(=\dfrac{1}{7}Σ\left(\dfrac{2}{x+3y}+\dfrac{1}{y+3z}+\dfrac{4}{z+3x}\right)\)

\(=\dfrac{1}{7}Σ\dfrac{7}{x+3y}=Σ\dfrac{1}{x+3y}=R.H.S\)

Áp dụng bất đẳng thức \(\dfrac{1}{x}+\dfrac{1}{y}\le\dfrac{4}{x+y}\) \(\forall x,y>0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+3y}+\dfrac{1}{y+2z+x}\le\dfrac{4}{2x+4y+2z}=\dfrac{2}{x+2y+z}\\\dfrac{1}{y+3z}+\dfrac{1}{z+2x+y}\le\dfrac{4}{2x+2y+4z}=\dfrac{2}{x+y+2z}\\\dfrac{1}{z+3x}+\dfrac{1}{x+2y+z}\le\dfrac{4}{4x+2y+2z}=\dfrac{2}{2x+y+z}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{x+3y}+\dfrac{1}{y+3z}+\dfrac{1}{z+3x}+\dfrac{1}{y+2z+x}+\dfrac{1}{z+2x+y}+\dfrac{1}{x+2y+z}\le\dfrac{2}{x+2y+z}+\dfrac{2}{x+y+2z}+\dfrac{2}{2x+y+z}\)

\(\Rightarrow VT\le\left(\dfrac{2}{x+2y+z}-\dfrac{1}{x+2y+z}\right)+\left(\dfrac{2}{x+y+2z}-\dfrac{1}{y+x+2z}\right)+\left(\dfrac{2}{2x+y+z}-\dfrac{1}{z+2x+y}\right)\)

\(\Rightarrow VT\le\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}+\dfrac{1}{2x+y+z}\)

\(\Leftrightarrow\dfrac{1}{x+3y}+\dfrac{1}{y+3z}+\dfrac{1}{z+3x}\le\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}+\dfrac{1}{2x+y+z}\) ( đpcm )

cau nay cau de y mot y la ra

chi lam the nay thoi cac cai sau cau dua vao ma lam tuong tu\(\dfrac{1}{x+3y}+\dfrac{1}{x+y+2z}\ge\dfrac{4}{2x+4y+2z}=\dfrac{2}{x+2y+z}\)

Áp dụng BĐT Cauchy-Schwarz:

\(\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{16}{3x+3y+2z}\\ \Leftrightarrow\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\\ \Leftrightarrow\sum\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\right)=\dfrac{4}{16}\cdot6=\dfrac{3}{2}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)

Ta có :

\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(2x+y+z\right)+\left(2y+x+z\right)}\)(1)

Áp dụng BĐT \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow\left(1\right)\le\dfrac{1}{4}\left(\dfrac{1}{x+y+x+z}+\dfrac{1}{y+x+y+z}\right)\le\dfrac{1}{4}\left(\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)\right)\)

\(=\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)

tương tự với hai ông còn lại sau đó cộng lại ta được:

\(\Sigma\dfrac{1}{3x+3y+2z}\le\dfrac{24}{16}=\dfrac{3}{2}\)

Bài 1:

\((x,y,z)=(\frac{2a^2}{bc}; \frac{2b^2}{ca}; \frac{2c^2}{ab})\) (\(a,b,c>0\) )

Khi đó:

\(\text{VT}=\frac{\frac{4a^4}{b^2c^2}}{\frac{4a^4}{b^2c^2}+\frac{4a^2}{bc}+1}+\frac{\frac{4b^4}{c^2a^2}}{\frac{4b^4}{c^2a^2}+\frac{4b^2}{ca}+4}+\frac{\frac{4c^4}{a^2b^2}}{\frac{4c^4}{a^2b^2}+\frac{4c^2}{ab}+4}\)

\(=\frac{a^4}{a^4+a^2bc+b^2c^2}+\frac{b^4}{b^4+b^2ac+a^2c^2}+\frac{c^4}{c^4+c^2ab+a^2b^2}\)

\(\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+a^2bc+b^2ac+c^2ab+(a^2b^2+b^2c^2+c^2a^2)}\)

(Áp dụng BĐT Cauchy_Schwarz)

Theo BĐT Cauchy dễ thấy:

\(a^2b^2+b^2c^2+c^2a^2\geq a^2bc+b^2ca+c^2ab\)

\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)}=\frac{(a^2+b^2+c^2)^2}{(a^2+b^2+c^2)^2}=1\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=2$

Bài 2:

Đặt \((x,y,z)=\left(\frac{a}{b};\frac{b}{c}; \frac{c}{a}\right)\)

Ta có:

\(\text{VT}=\left(\frac{a}{b}+\frac{c}{b}-1\right)\left(\frac{b}{c}+\frac{a}{c}-1\right)\left(\frac{c}{a}+\frac{b}{a}-1\right)\)

\(=\frac{(a+c-b)(b+a-c)(c+b-a)}{abc}\)

Áp dụng BĐT Cauchy:

\((a+c-b)(b+a-c)\leq \left(\frac{a+c-b+b+a-c}{2}\right)^2=a^2\)

\((b+a-c)(c+b-a)\leq \left(\frac{b+a-c+c+b-a}{2}\right)^2=b^2\)

\((a+c-b)(c+b-a)\leq \left(\frac{a+c-b+c+b-a}{2}\right)^2=c^2\)

Nhân theo vế:

\(\Rightarrow [(a+c-b)(b+a-c)(c+b-a)]^2\leq (abc)^2\)

\(\Rightarrow (a+c-b)(b+a-c)(c+b-a)\leq abc\)

\(\Rightarrow \text{VT}\leq 1\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=1$

\(\)

2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)

1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).

CM:....

Đặt 2x = x', 2z = z'.

Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)

\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)

\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)

\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)

\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)

\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)