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

Áp dụng Bđt Bunhia

\(\left(x+2y\right)^2\le\left(1^2+2^2\right)\left(x^2+y^2\right)\le5\cdot2\left(x+2y\right)\)

\(\Rightarrow x+2y\le10\)

Dpcm

Ta có: \(\left\{{}\begin{matrix}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{matrix}\right.\)

\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)

\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)

\(\Rightarrow x=y=z=-1\)(do \(\left(x+1\right)^2,\left(y+1\right)^2,\left(z+1\right)^2\ge0\forall x,y,z\))

a) \(A=x^{2020}+y^{2020}+z^{2020}=\left(-1\right)^{2020}+\left(-1\right)^{2020}+\left(-1\right)^{2020}=1+1+1=3\)

b) \(B=\dfrac{1}{x^{2020}}+\dfrac{1}{y^{2020}}+\dfrac{1}{z^{2020}}=\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}=\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}=3\)

bạn nhóm thành các bình phương nhé. còn dư 4xy với 1.

bạn trình bày cho mình đc ko?

Ta có:  2 x 2 + 1 2 ≥ 2 x ;  2 y 2 + 1 2 ≥ 2 y và  x 2 + y 2 ≥ 2 x y

Cộng vế với vế các BĐT trên ta được:

3 x 2 + y 2 + 1 ≥ 2 x + y + x y = 5 2

=> A =  x 2 + y 2 ≥ 1 2

Từ đó tìm được  A m i n = 1 2 <=> x = y =  1 2

\(VT=6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

\(=6\left(x+y+z\right)^2-2\left(xy+yz+xz\right)+2\frac{9}{2x+y+z+x+2y+z+x+y+2z}\)

\(\ge6\left(x+y+z\right)^2-2\frac{\left(x+y+z\right)^2}{3}+2\frac{9}{4\left(x+y+z\right)}\)

\(=\: 6\cdot\left(\frac{3}{4}\right)^2-2\cdot\frac{\left(\frac{3}{4}\right)^2}{3}+2\cdot\frac{9}{4\cdot\frac{3}{4}}=9\)

Do \(x^2+y^2=1\Rightarrow-1\le x;y\le1\Rightarrow\left\{{}\begin{matrix}y+1\ge0\\1-y\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y^2\left(y+1\right)\ge0\\y^2\left(1-y\right)\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y^3\ge-y^2\\y^3\le y^2\end{matrix}\right.\)

Với mọi số thực x ta có:

\(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left(x-1\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x\ge-x^2-1\\2x\le x^2+1\end{matrix}\right.\)

Do đó: \(\left\{{}\begin{matrix}P=2x+y^3\ge-x^2-1-y^2=-2\\P=2x+y^3\le x^2+1+y^2=2\end{matrix}\right.\)

\(P_{min}=-2\) khi \(\left(x;y\right)=\left(-1;0\right)\)

\(P_{max}=2\) khi \(\left(x;y\right)=\left(1;0\right)\)

Theo gt: x² +y² ≤ 2 (x + 2y) x² + y² ≤ 2(x + 2y)

Ta có: (x + 2y)² ≤ (1² + 2²)(x² + y²) ≤ 5.2(x + 2y)(x + 2y)² ≤ (1² + 2²)(x²+y²) ≤ 5.2(x+2y)

⇒ x + 2y ≤ 10 ⇒ x + 2y ≤ 10 (đpcm)