Câu hỏi của bach nhac lam

Question

1. Cho $x,y,z\in\left(0,1\right)$ và $xyz=\left(1-x\right)\left(1-y\right)\left(1-z\right)$. Cmr: $x^2+y^2+z^2\ge\frac{3}{4}$

2. \(\left\{{}\begin{matrix}x,y,z\ge0\x^2+y^2+z^2+xyz=4\end{matrix...

Nguyễn Việt Lâm · Accepted Answer

Câu 1:

$2xyz=1-\left(x+y+z\right)+xy+yz+zx$

$\Rightarrow xy+yz+zx=2xyz+\left(x+y+z\right)-1$

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

$=\left(x+y+z\right)^2-2\left(x+y+z\right)-4xyz+2$

$VT\ge\left(x+y+z\right)^2-2\left(x+y+z\right)-\frac{4}{27}\left(x+y+z\right)^3+2$

$VT\ge\frac{4}{27}\left[\frac{15}{4}-\left(x+y+z\right)\right]\left(x+y+z-\frac{3}{2}\right)^2+\frac{3}{2}\ge\frac{3}{2}$

(Do $0< x;y;z< 1\Rightarrow x+y+z< 3< \frac{15}{4}$)

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

$2xyz=1-\left(x+y+z\right)+xy+yz+zx$

$\Rightarrow xy+yz+zx=2xyz+\left(x+y+z\right)-1$

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

$=\left(x+y+z\right)^2-2\left(x+y+z\right)-4xyz+2$

$VT\ge\left(x+y+z\right)^2-2\left(x+y+z\right)-\frac{4}{27}\left(x+y+z\right)^3+2$

$VT\ge\frac{4}{27}\left[\frac{15}{4}-\left(x+y+z\right)\right]\left(x+y+z-\frac{3}{2}\right)^2+\frac{3}{2}\ge\frac{3}{2}$

(Do $0< x;y;z< 1\Rightarrow x+y+z< 3< \frac{15}{4}$)

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

Nguyễn Việt Lâm · Answer

Câu 2:

Từ điều kiện bài này có thể đặt ẩn phụ và AM-GM ra luôn kết quả, nhưng hơi rắc rối khi người ta hỏi từ đâu mà có cách đặt ẩn phụ như vậy, do đó ta giải trâu :D

$x^2+y^2+z^2+xyz=4$

$\Leftrightarrow\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{4}+2\left(\frac{x}{2}.\frac{y}{z}.\frac{z}{2}\right)=1$

$\Leftrightarrow\frac{xy}{2z}.\frac{xz}{2y}+\frac{xy}{2z}.\frac{yz}{2x}+\frac{yz}{2x}.\frac{xz}{2y}+2\left(\frac{xy}{2z}.\frac{yz}{2x}.\frac{xy}{2y}\right)=1$

Đặt $\left(\frac{xy}{2z};\frac{zx}{2y};\frac{yz}{2x}\right)=\left(m;n;p\right)\Rightarrow mn+np+pn+2mnp=1$

$\Leftrightarrow2\left(n+1\right)\left(m+1\right)\left(p+1\right)=\left(n+1\right)\left(m+1\right)+\left(n+1\right)\left(p+1\right)+\left(m+1\right)\left(p+1\right)$

$\Leftrightarrow\frac{1}{n+1}+\frac{1}{m+1}+\frac{1}{p+1}=2$

$\Leftrightarrow1=\frac{n}{n+1}+\frac{m}{m+1}+\frac{p}{p+1}\ge\frac{\left(\sqrt{n}+\sqrt{m}+\sqrt{p}\right)^2}{m+n+p+3}$

$\Leftrightarrow m+m+p+2\left(\sqrt{mn}+\sqrt{np}+\sqrt{mp}\right)\le m+n+p+3$

$\Leftrightarrow\sqrt{mn}+\sqrt{np}+\sqrt{mp}\le\frac{3}{2}$

$\Leftrightarrow\frac{x}{2}+\frac{y}{2}+\frac{z}{2}\le\frac{3}{2}\Leftrightarrow x+y+z\le3$Câu trả lời: Câu 2: