\(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{\left(x-y\right)^2}{xy}\ge0\)

\(x^2+y^2-z^2>0\Rightarrow x^2+2xy+y^2-z^2>0\)

\(\Rightarrow\left(x+y\right)^2-z^2>0\)

\(\Rightarrow\left(x+y-z\right)\left(x+y+z\right)>0\)

Mà x;y;z>0 \(\Rightarrow x+y+z>0\)

\(\Rightarrow x+y-z>0\)

bạn cảm ơn ai vay có bn ấy có giup bn làm đau

mk chua hok den nen ko co bit lam

cảm ơn b nhé

Mình đã làm được rồi

b1:

x-y=5->x=y+5

->x-3y/5-2y=y+5-3y/5-2y=5-2y5-2y=1

->đpcm

Ta có: \(\frac{x^3+y^3+z^3-3xyz}{x+y+z}\)

\(=\frac{\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz}{x+y+z}\)

\(=\frac{\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)}{x+y+z}\)

\(=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-yz-zx-3xy\right)}{x+y+z}\)

\(=x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\ge0\left(\forall x,y,z\right)\)

=> đpcm

Đặt \(P=\left(\dfrac{x-y}{z}+\dfrac{y-z}{x}+\dfrac{z-x}{y}\right)\left(\dfrac{z}{x-y}+\dfrac{x}{y-z}+\dfrac{y}{z-x}\right)=9\)

Đặt \(\left\{{}\begin{matrix}\dfrac{x-y}{z}=a\\\dfrac{y-z}{x}=b\\\dfrac{x-z}{y}=c\end{matrix}\right.\)

\(\Leftrightarrow P=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\\ =1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\\ =3+\dfrac{a+c}{b}+\dfrac{a+b}{c}+\dfrac{b+c}{a}\)

Ta có \(\dfrac{a+c}{b}=\dfrac{\dfrac{x-y}{z}+\dfrac{z-x}{y}}{\dfrac{y-z}{x}}=\dfrac{xy-y^2+z^2-xz}{yz}\cdot\dfrac{x}{y-z}\)

\(=\dfrac{\left(z-y\right)\left(y+z-x\right)x}{yz\left(y-z\right)}=\dfrac{x\left(x-y-z\right)}{yz}\)

Mà \(x+y+z=0\Leftrightarrow x=-y-z\)

\(\Leftrightarrow\dfrac{a+c}{b}=\dfrac{x\left(x+x\right)}{yz}=\dfrac{2x^2}{yz}\)

Cmtt ta được \(\dfrac{a+b}{c}=\dfrac{2y^2}{xz};\dfrac{b+c}{a}=\dfrac{2z^2}{xy}\)

Cộng vế theo vế

\(\Leftrightarrow P=\dfrac{2x^2}{yz}+\dfrac{2y^2}{xz}+\dfrac{2z^2}{xy}+3=\dfrac{2x^3+2y^3+2z^3}{xyz}+3\\ \Leftrightarrow P=\dfrac{2\left(x^3+y^3+z^3\right)}{xyz}+3\)

Lại có \(x+y+z=0\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\\ \Leftrightarrow x^3+y^3+z^3=3xyz\)

Thế vào \(P\)

\(\Leftrightarrow P=\dfrac{2\cdot3xyz}{xyz}+3=6+3=9\)

BĐT cần chứng minh tương đương:

\(x^4+y^4\ge x^3y+xy^3\)

\(\Leftrightarrow x^4-x^3y+y^4-xy^3\ge0\)

\(\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)\left(x^3-y^3\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

Ta có bất đẳng thức $a^2+b^2 \geq \dfrac{(a+b)^2}{2}

$⇔2.(a^2+b^2) \geq (a+b)^2$

$⇔(a-b)^2 \geq 0$ (đúng)

Áp dụng bất đẳng thức trên cho $\dfrac{x}{y}$ và $\dfrac{y}{x}$ có:

$\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2} $

$\geq \dfrac{(\dfrac{x}{y}+\dfrac{y}{x})^2}{2}$

Hay $2.\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2} \geq (\dfrac{x}{y}+\dfrac{y}{x})^2$

Áp dụng bất đẳng thức Cauchy (Cô-si) có:

$\dfrac{x}{y}+\dfrac{y}{x} \geq 2.\sqrt[]{\dfrac{x}{y}.\dfrac{y}{x}}=2$

Nên $(\dfrac{x}{y}+\dfrac{y}{x}).(\dfrac{x}{y}+\dfrac{y}{x}) \geq 2.(\dfrac{x}{y}+\dfrac{y}{x})$

Hay $ (\dfrac{x}{y}+\dfrac{y}{x})^2  \geq 2.(\dfrac{x}{y}+\dfrac{y}{x})$

Suy ra $2.\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2} \geq 2.(\dfrac{x}{y}+\dfrac{y}{x})$

Hay $\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2} \geq (\dfrac{x}{y}+\dfrac{y}{x})(đpcm)$

Dấu $=$ xảy ra $⇔x=y$

Thiếu đề. Nếu x = 1; y = 1 thì không thỏa mãn đề bài

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