M+2019=2xy−yz−zx+2020M+2019=2xy−yz−zx+2020

=2xy−yz−zx+x2+y2+z2=2xy−yz−zx+x2+y2+z2

=(x+y−z2)2+3z24≥0=(x+y−z2)2+3z24≥0

⇒Mmin=0⇒Mmin=0 khi ⎧⎩⎨⎪⎪⎪⎪x+y−z2=03z24=0x2+y2+z2=2020{x+y−z2=03z24=0x2+y2+z2=2020

⇔⎧⎩⎨⎪⎪x+y=0z=0x2+y2=2020⇔{x+y=0z=0x2+y2=2020 ⇒⎧⎩⎨⎪⎪x=±1010−−−−√y=−xz=0

mình không hiểu ạ

x² + y² + z² = xy + yz + zx

⇔ 2( x² + y² + z² ) = 2( xy + yz + zx )

⇔ 2x² + 2y² + 2z² = 2xy + 2yz + 2zx

⇔ 2x² + 2y² + 2z² = 2xy + 2yz + 2zx

⇔ 2x² + 2y² + 2z² - 2xy - 2yz - 2zx = 0

⇔ ( x² - 2xy + y² ) + ( y² - 2yz + z² ) + ( z² - 2xz + x² ) = 0

⇔ ( x - y )² + ( y - z )² + ( z - x )² = 0

Vì : \(\hept{\begin{cases}\left(x-y\right)^2\\\left(y-z\right)^2\\\left(z-x\right)^2\end{cases}}\ge0\forall x,y,z\)=> ( x - y )² + ( y - z )² + ( z - x )² ≥ 0 ∀ x, y, z

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\Leftrightarrow x=y=z\)

Khi đó M = ( x - y + 1 )²⁰¹⁹ + ( y - z + 1 )²⁰²⁰ < đã sửa >

               = ( x - x + 1 )²⁰¹⁹ + ( y - y + 1 )²⁰²⁰

               = 1²⁰¹⁹ + 1²⁰²⁰

               = 1 + 1 = 2

\(A=\dfrac{xyz.x}{xy+xyz.x+xyz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{xyz+yz+y}\)

\(=\dfrac{xz}{1+xz+z}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)

\(=\dfrac{xyz}{y+xyz+yz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)

\(=\dfrac{2019}{y+2019+yz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)

\(=\dfrac{yz+y+2019}{yz+y+2019}=1\)

\(T\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+x+y+z}=\dfrac{x+y+z}{2}\ge\dfrac{2019}{2}\)

áp dụng BĐT:\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\) với a,b,c,x,y,z là số dương

ta có BĐT Bunhiacopxki cho 3 bộ số:\(\left(\dfrac{a}{\sqrt{x}};\sqrt{x}\right);\left(\dfrac{b}{\sqrt{y}};\sqrt{y}\right);\left(\dfrac{c}{\sqrt{z}};\sqrt{z}\right)\)

ta có :

\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\left(x+y+z\right)\)\(=\left[\left(\dfrac{a}{\sqrt{x}}\right)^2+\left(\dfrac{b}{\sqrt{y}}\right)^2+\left(\dfrac{c}{\sqrt{z}}\right)^2\right]\).\(\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\right]\)\(\ge\left(\dfrac{a}{\sqrt{x}}.\sqrt{x}+\dfrac{b}{\sqrt{y}}.\sqrt{y}+\dfrac{c}{\sqrt{z}}.\sqrt{z}\right)^2=\left(a+b+c\right)^2\)

lúc đó ta có :\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}\)

ta có \(T=\dfrac{x^2}{x+\sqrt{yz}}+\dfrac{y^2}{y+\sqrt{zx}}+\dfrac{z^2}{z+\sqrt{xy}}\)\(\ge\dfrac{\left(x+y+z\right)^2}{x+\sqrt{yz}+y+\sqrt{zx}+z+\sqrt{xy}}\) mà ta có :

\(\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\)\(\le\dfrac{x+y}{2}+\dfrac{x+z}{2}+\dfrac{z+y}{2}\)\(\Rightarrow\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\le x+y+z\)

\(\Rightarrow T=\dfrac{2019}{2}\Leftrightarrow x=y=z=673\)

vậy \(\text{MinT}=\dfrac{2019}{2}\) khi và chỉ khi x=y=z=673

Ta có:

\(\sqrt{2019x+yz}=\sqrt{x\left(x+y+z\right)+yz}\)\(=\sqrt{x^2+xy+xz+yz}=\sqrt{x^2+yz+x\left(y+z\right)}\)

Áp dụng BĐT AM-GM cho các số không âm, ta có:

\(x^2+yz\ge2x\sqrt{yz}\)

\(\Rightarrow x^2+yz+x\left(y+z\right)\ge x\left(y+z+2\sqrt{yz}\right)\)

\(\Leftrightarrow2019x+yz\ge\left[\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)\right]^2\)\(\ge0\)

\(\Rightarrow\sqrt{2019x+yz}\ge\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)\)

\(\Rightarrow x+\sqrt{2019x+yz}\ge\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

\(\Rightarrow\frac{x}{x+\sqrt{2019x+yz}}\le\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

CMTT, ta có:

\(\frac{y}{y+\sqrt{2019y+zx}}\le\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\),\(\frac{z}{z+\sqrt{2019z+xy}}\le\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

\(\Rightarrow M\le\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)

\(''=''\Leftrightarrow x=y=z=673\)

Đề là \(M=\sum\frac{x}{x+\sqrt{2019+yz}}\) hay \(M=\sum\frac{x}{x+\sqrt{2019x+yz}}\) bạn?

Nếu là đề bạn đúng thì mình bó tay.

\(P=\frac{2019xz}{xyz+2019xz+2019z}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)

\(=\frac{2019xz}{2019+2019xz+2019z}+\frac{y}{y\left(xz+z+1\right)}+\frac{z}{xz+z+1}\)

\(\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}=1\)

Đk: $x\geq \frac{1}{2}$

Pt $\Leftrightarrow 4x^2+3x-7=4(\sqrt{x^3+3x^2}-2)+2(\sqrt{2x-1}-1)$

$\Leftrightarrow +4\frac{(x-1)(x+2)^2}{\sqrt{x^3+3x^2}+2}+4\frac{x-1}{\sqrt{2x-1}+1}-(x-1)(4x+7)=0$

$\Leftrightarrow (x-1)[\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-(4x+7)]=0$

$\Leftrightarrow x=1\vee \frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7=0$ $(*)$

Xét hàm số $f(x)=\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7,x\in [\frac{1}{2};+\infty )$ thì $f(x)>0,\forall x\in [\frac{1}{2};+\infty )$

$\Rightarrow $ Pt $(*)$ vô nghiệm

Ta có : \(A=\frac{2019}{x+xy+1}+\frac{2019}{y+yz+1}+\frac{2019}{z+zx+1}=2019\left(\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+zx+1}\right)\)

\(=2019\left(\frac{z}{xz+xyz+z}+\frac{xz}{xyz+xyz^2+xz}+\frac{1}{z+zx+1}\right)\)

\(=2019\left(\frac{z}{xz+z+1}+\frac{xz}{1+z+xz}+\frac{1}{z+zx+1}\right)\)(vì xyz = 1)

\(=2019\left(\frac{z+xz+1}{xz+z+1}\right)=2019\)

Vậy A = 2019

\(x^{2019}+y^{2019}+z^{2019}=\left(x+y+z\right)^{2019}\)

Em xin lỗi, đây mới là đề đúng ạ !!