(x+y)(y+z)(x+z)=8xyz

<=>\((xy+xz+y^2+yz)(x+z)=8xyz\)

<=>\(x^2y+x^2z+y^2z+xyz+xyz+xz^2+z^2y+yz^2=8xyz\)

<=> \(x^2y+x^2z+y^2x+xz^2+y^2z+yz^2-6xyz=0\)

<=> \(y(x^2+z^2-2xz)+x(y^2-2yz+z^2)+z(y^2-2yx+x^2)=0\)

<=>\(y(x-z)^2+x(y-z)^2+z(x-y)^2=0\)

Mà x,y,z dương

=> \((x-z)^2=0=>x=z\)

\((x-y)^2=0=>x=y\)

\((y-z)^2=0=>y=z\)

Vậy x=y=z

cảm ơn ạ

=>

=>

=>

Tương tự, ta có: 

Do đó, ta có:

(ĐPCM)

a) x > 0

b) y ≥ 0

c) ∀α ∈ R, |α| ≥ 0

d) ∀a, b > 0, 

\(\frac{a}{c}+\frac{c}{a}-2=\frac{a}{c}+\frac{c}{a}-2\cdot\sqrt{\frac{a}{c}}\cdot\sqrt{\frac{c}{a}}=\left(\sqrt{\frac{a}{c}}-\sqrt{\frac{c}{a}}\right)^2\ge0\forall a,c\) >0

=>\(\frac{a}{c}+\frac{c}{a}\ge2\left(1\right)\)

\(\frac{a}{b}+\frac{b}{a}-2=\frac{a}{b}-2\cdot\sqrt{\frac{a}{b}}\cdot\sqrt{\frac{b}{a}}+\frac{b}{a}=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\ge0\forall a,b>0\)

=>\(\frac{a}{b}+\frac{b}{a}\ge2\left(2\right)\)

\(\frac{b}{c}+\frac{c}{b}-2=\frac{b}{c}-2\cdot\sqrt{\frac{b}{c}}\cdot\sqrt{\frac{c}{b}}+\frac{c}{b}=\left(\sqrt{\frac{b}{c}}-\sqrt{\frac{c}{b}}\right)^2\ge0\forall b,c>0\)

=>\(\frac{b}{c}+\frac{c}{b}\ge2\left(3\right)\)

Từ (1),(2),(3) suy ra \(\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge2+2+2\)

=>\(\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\forall a,b,c>0\)

=>ĐPCM

Điều kiện \(a>0\)

\(A=\sqrt[4]{\frac{3}{4a}}.\sqrt[4]{\frac{4a}{3}}.x\sqrt{a-x^4}\le\sqrt[4]{\frac{3}{4a}}\left(-x^4+\sqrt{\frac{4a}{3}}x^2+a\right)\)

\(A\le\sqrt[4]{\frac{3}{4a}}\left[\frac{4a}{3}-\left(x^2-\sqrt{\frac{a}{3}}\right)^2\right]\le\frac{4a}{3}\sqrt[4]{\frac{3}{4a}}\)

Dấu "=" xảy ra khi \(x=\sqrt[4]{\frac{a}{3}}\)

\(A^2=\left(x-y\right)^2=\left(1.x+\dfrac{1}{2}.\left(-2y\right)\right)^2\le\left(1+\dfrac{1}{4}\right)\left(x^2+4y^2\right)=\dfrac{5}{4}\)

\(\Rightarrow A\le\dfrac{\sqrt{5}}{2}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(-\dfrac{2\sqrt{5}}{5};\dfrac{\sqrt{5}}{10}\right);\left(\dfrac{2\sqrt{5}}{5};-\dfrac{\sqrt{5}}{10}\right)\)