Đặt \(A=x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)

\(\Leftrightarrow A=x+y+z+\dfrac{9}{9x}+\dfrac{9}{9y}+\dfrac{9}{9z}\)

\(\Leftrightarrow A=x+y+z+\dfrac{1}{9x}+\dfrac{8}{9x}+\dfrac{1}{9y}+\dfrac{8}{9y}+\dfrac{1}{9z}+\dfrac{8}{9z}\)

\(\Leftrightarrow A=\left(x+\dfrac{1}{9x}\right)+\left(y+\dfrac{1}{9y}\right)+\left(z+\dfrac{1}{9z}\right)+\left(\dfrac{8}{9x}+\dfrac{8}{9y}+\dfrac{8}{9z}\right)\)

\(\Leftrightarrow A=\left(x+\dfrac{1}{9x}\right)+\left(y+\dfrac{1}{9y}\right)+\left(z+\dfrac{1}{9z}\right)+\dfrac{8}{9}.\left(\dfrac{1^2}{x}+\dfrac{1^2}{y}+\dfrac{1^2}{z}\right)\)

\(\Rightarrow A\ge2\sqrt{x.\dfrac{1}{9x}}+2\sqrt{y.\dfrac{1}{9y}}+2\sqrt{z.\dfrac{1}{9z}}+\dfrac{8}{9}.\dfrac{\left(1+1+1\right)^2}{x+y+z}\)

\(\Rightarrow A\ge2\sqrt{\dfrac{1}{9}}+2\sqrt{\dfrac{1}{9}}+2\sqrt{\dfrac{1}{9}}+\dfrac{8}{9}.\dfrac{3^2}{1}\)

\(\Rightarrow A\ge2.\dfrac{1}{3}.3+8=2+8=10\)

Vậy ta có BĐT cần chứng minh.

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

\(A=\dfrac{2\left(x^3+y^3\right)}{\left(x^4+y^2\right)\left(x^2+y^4\right)}=2.\dfrac{\left(x^3+y^3\right)}{x^4y^4+x^2y^2+x^6+y^6}\)

\(=2.\dfrac{\left(x^3+y^3\right)}{1+1+x^6+y^6}=2.\dfrac{x^3+y^3}{x^6+y^6+2x^3y^3}=2.\dfrac{x^3+y^3}{\left(x^3+y^3\right)^2}=\dfrac{2}{x^3+y^3}\left(1\right)\)

Áp dụng bất đẳng thức Cauchy ta có:

\(x^3+y^3+1\ge3\sqrt{xy.1}=3\)

\(\Rightarrow x^3+y^3\ge2\Rightarrow\dfrac{2}{x^3+y^3}\le1\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow A\le1\)

Dấu "=" xảy ra khi x=y=1.

Vậy MaxA là 1, đạt được khi x=y=1.

Lời giải:
Áp dụng BĐT AM-GM:
$1=xy+yz+xz+2xyz\leq \frac{(x+y+z)^2}{3}+2.\frac{(x+y+z)^3}{27}$

$\Leftrightarrow 1\leq \frac{t^2}{3}+\frac{2t^3}{27}$ (đặt $x+y+z=t$)

$\Leftrightarrow 2t^3+9t^2-27\geq 0$

$\Leftrightarrow (t+3)^2(2t-3)\geq 0$

$\Leftrightarrow 2t-3\geq 0$
$\Leftrightarrow t\geq \frac{3}{2}$ hay $x+y+z\geq \frac{3}{2}$ (đpcm)

Dấu "=" xảy ra khi $x=y=z=\frac{1}{2}$

Lời giải:

$A=(x+y)(x^2-xy+y^2)+x^2+y^2=2(x^2-xy+y^2)+x^2+y^2=2(x^2+y^2)+(x-y)^2$

$\geq 2(x^2+y^2)=(1^2+1^2)(x^2+y^2)\geq (x+y)^2=2^2=4$ (theo BĐT Bunhiacopxky)

Vậy $A_{\min}=4$. Giá trị này đạt tại $x=y=1$

A=(x+y)³ - 3xy(x+y)+3x²y²

=8-6xy+3x²y²

=3(x²y²-2xy+1)+5

=3(xy+1)²+5 ≥5

dấu = xảy ra ⇔ xy=1 ⇒ x=y=1

Ta có:

Đặt \(A=x+y+\dfrac{1}{x}+\dfrac{1}{y}\)

\(\Leftrightarrow A=x+y+\dfrac{4}{4x}+\dfrac{4}{4y}\)

\(\Leftrightarrow A=x+y+\dfrac{1}{4x}+\dfrac{3}{4x}+\dfrac{1}{4y}+\dfrac{3}{4y}\)

\(\Leftrightarrow A=\left(x+\dfrac{1}{4x}\right)+\left(y+\dfrac{1}{4y}\right)+\left(\dfrac{3}{4x}+\dfrac{3}{4y}\right)\)

\(\Rightarrow A\ge2\sqrt{x.\dfrac{1}{4x}}+2\sqrt{y.\dfrac{1}{4y}}+\dfrac{3}{4}.\dfrac{4}{x+y}\)

\(\ge2.\sqrt{\dfrac{1}{4}}+2\sqrt{\dfrac{1}{4}}+\dfrac{3}{4}.\dfrac{4}{1}\)

\(=2.\dfrac{1}{2}+2.\dfrac{1}{2}+3=1+1+3=5\)

Vậy ta có đpcm. Dấu"=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4x}\\y=\dfrac{1}{4y}\\x=y\\x+y=1\end{matrix}\right.\) \(\Leftrightarrow x=y=\dfrac{1}{2}\left(tm\right)\)

Ta có \(x^2+y^3\ge x^3+y^4\Leftrightarrow x^2+y^2+y^3\ge x^3+y^2+y^4\)

Áp dụng bđt AM-GM ta có \(y^4+y^2\ge2y^3\)

\(\Rightarrow x^2+y^3+y^2\ge x^3+2y^3\)

\(\Rightarrow x^3+y^3\le x^2+y^2\left(1\right)\)

Áp dụng bđt Cauchy - Schwarz ta có 

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

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

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

Lại có

\(\left(x+y\right)^2\le2\left(x^2+y^2\right)\le2\left(x+y\right)\)

\(\Rightarrow x+y\le2\left(3\right)\)

Từ (1),(2),(3) => đpcm

Đối với bài này ta cũng có thể chia các khoảng giá trị để chứng minh 

(Nhưng hơi dài và khó hiểu nên mình k làm ) 

Học tốt!!!!!!!!!

Dấu "=" xảy ra khi x=y=1