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Dự đoán đẳng thức xảy ra khi x = y = z = 1.

Đặt x = 1 + a ; y = 1 + b , ( a , b $\in$∈ R ). Từ giả thiết suy ra z = 1 - a - b.

Ta có: 
$x^2+y^2+z^2+xy+yz+zx$x²+y²+z²+xy+yz+zx$=\left(1+a\right)^2+\left(1+b\right)^2+\left(1-a-b\right)^2+\left(1+a\right)\left(1+b\right)+\left(1+b\right)\left(1-a-b\right)+\left(1-a-b\right)\left(1+a\right)=\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}+6\ge6.$=(1+a)²+(1+b)²+(1−a−b)²+(1+a)(1+b)+(1+b)(1−a−b)+(1−a−b)(1+a)=(a+b2 )²+3b²4 +6≥6.

Đẳng thức xảy ra khi và chỉ khi.

$b=0;a+\frac{b}{2}=0\Leftrightarrow a=0;b=0\Leftrightarrow x=y=z=1.$b=0;a+b2 =0⇔a=0;b=0⇔x=y=z=1.

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Ta có: 

\(x^3+y^3+z^3-3xyz\)

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

\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)

\(=\left(x+y+z\right)^3-3\left(x+y+z\right)\left(x+y\right).z-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yx-3xz-3yz-3xy\right)\)

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

=> \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz\)

(x+y+z)² = x² + y² + z² + 2(xy +yz +zx)

2) \(43^{2020}+43^{2021}=43^{2020}\left(1+43\right)=43^{2020}.44\)

Mà \(44⋮11\Rightarrow43^{2020}.44⋮11\Rightarrow43^{2020}+43^{2021}⋮11\)

Phần 1 đang nghĩ -.-

Sửa đề \(\left(x+y+z\right)^2-x^2-y^2-z^2=2\left(xy+yz+zx\right)\)

Ta có : \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\)(hằng đẳng thức cho  3 số )

\(\Rightarrow\left(x+y+z\right)^2-x^2-y^2-z^2=2\left(xy+yz+zx\right)\left(đpcm\right)\)

Vậy

Lời giải:

Ta có:

\(3=xy+yz+xz\leq \frac{(x+y+z)^2}{3}\Rightarrow x+y+z\geq 3\)

Áp dụng BĐT AM-GM:

\(x^3+8=(x+2)(x^2-2x+4)\leq \left(\frac{x+2+x^2-2x+4}{2}\right)^2\)

\(\Rightarrow \sqrt{x^3+8}\leq \frac{x^2-x+6}{2}\Rightarrow \frac{x^2}{\sqrt{x^3+8}}\geq \frac{2x^2}{x^2-x+6}\)

Thực hiện tương tự với các phân thức còn lại và cộng theo vế:

\(\Rightarrow \text{VT}\geq \underbrace{2\left(\frac{x^2}{x^2-x+6}+\frac{y^2}{y^2-y+6}+\frac{z^2}{z^2-z+6}\right)}_{M}\)

Áp dụng BĐT Cauchy-Schwarz:

\(M\geq \frac{2(x+y+z)^2}{x^2-x+6+y^2-y+6+z^2-z+6}=\frac{2(x+y+z)^2}{x^2+y^2+z^2-(x+y+z)+18}\)

\(\Leftrightarrow M\geq \frac{2(x+y+z)^2}{(x+y+z)^2-(x+y+z)+12}\) (do $xy+yz+xz=3$)

Mà :

\(\frac{(x+y+z)^2}{(x+y+z)^2-(x+y+z)+12}-1=\frac{(x+y+z)^2+(x+y+z)-12}{(x+y+z)^2-(x+y+z)+12}=\frac{(x+y+z-3)(x+y+z+4)}{(x+y+z)^2-(x+y+z)+12}\geq 0\) do $x+y+z\geq 0$

Do đó: \(M\geq 1\Rightarrow \text{VT}\geq 1\) (đpcm)

Dấu bằng xảy ra khi \(x=y=z=1\)

Ta có:

VT= \(\left(x+y+z\right)^2-x^2-y^2-z^2\)

\(=x^2+y^2+z^2+2xy+2yz+2zx-x^2-y^2-z^2\)

\(=2\left(xy+yz+zx\right)\) = VP

=> đpcm

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

Biến đổi vế trái:

VT\(\)\(\)\(=\left[\left(x+y\right)+z\right]^2-x^2-y^2-z^2\)

\(=\left(x+y\right)^2+2\left(x+y\right)z+z^2-x^2-y^2-z^2\)

\(=x^2+2xy+y^2+2xz+2yz+z^2-x^2-y^2-z^2\)\

\(=2xy+2yz+2zx\)

\(=2\left(xy+yz+zx\right)=\) VP

Ta có:

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

\(=x^2+y^2+z^2+2xy+2yz+2zx-x^2-y^2-z^2\)

\(=2xy+2yz+2zx\)

\(=2\left(xy+yz+zx\right)\)

Có: \(\left(x+y+z\right)^2-x^2-y^2-z^2\) 

\(=x^2+y^2+z^2+2xy+2yz+2xz-x^2-y^2-z^2\)

\(=2xy+2yz+2xz\)

\(=2\left(xy+yz+xz\right)\)

\(\left[\left(x+y\right)+z\right]^2=\left[\left(x+y\right)^2+2.\left(x+y\right)z+z^2\right]=x^2+2xy+y^2+2xz+2yz+z^2\)\(+z^2\)

Thay vào: x^2+y^2+z^2+ 2xy+2yz+2xz - x^2 - y^2 - z^2= 2(xy+yz+xz) (đpcm)