Bài 100915

Question

Cho $x,y,z\geq 0$
$1/$Hãy chứng minh:$xyz\geq \left ( y+z-x \right )\left (z+x-y \right )\left ( x+y-z \right )$
$2/$Nếu thêm điều kiện: $x+y+z=1$,chứng minh:
$0\leq xy+yz+zx-2xyz\leq \frac{7}{27}$

score 1 · Answer 1

$1/$Không giảm tính tổng quát, ta xem:
$x\geq y\geq z$, vì vậy:$ \begin{cases}z+x-y\geq 0 \\ x+y-z\geq 0 \end{cases}$
Nếu: $y+z-x<0$: BĐT luôn đúng.
Nếu: $y+z-x\geq 0$: lúc đó:
$\begin{cases}x^{2}\geq x^{2}-\left ( y-z \right )^{2}=\left ( x+y-z \right )\left ( x-y+z \right )\geq 0\\ y^{2}\geq y^{2}-\left ( z-x \right )^{2}=\left ( x+y-z \right )\left ( y+z-x \right )\geq 0 \\z^{2}\geq z^{2}-\left ( x-y \right )^{2}=\left ( x-y+z \right )\left (y+z-x \right )\geq 0\end{cases}$
Cho nên: $\left ( x yz\right )^{2}\geq [\left ( -x+y+z \right )\left ( x-y+z \right )\left ( x+y-z \right )]^{2}$
hay $xyz\geq\left ( -x+y+z \right )\left ( x-y+z \right )\left ( x+y-z \right ) $
$2/$Cũng vậy, ta vẫn xem: $x\geq y\geq z\geq 0$
và $x+y+z=1\Rightarrow 1\geq x\geq \frac{1}{3}\geq z\geq 0$
Lúc đó:
$xy+yz+zx-2xyz=xy\left (1-z \right )+yz\left (1-x \right )+zx\geq 0$
$4xy=\left ( x+y\right )^{2}-\left ( x-y\right )^{2}\leq \left ( x+y\right )^{2}=\left ( 1-z\right )^{2}$
$\Rightarrow xy+z+zx-2xyz=z\left ( x+y\right )+xy\left ( 1-2z\right )$
$\leq z\left ( 1-z\right )+\frac{1}{4}\left ( 1-z\right )^{2}\left ( 1-2z\right )= \frac{1}{4}\left ( 1+z^{2}-2z^{3} \right ) $
( vì $ 0\leq z\leq \frac{1}{3}$)
$\leq \frac{7}{27}-\frac{1}{4}\left ( z-\frac{1}{3} \right )^{2}\left ( 2z+\frac{1}{3} \right )\leq \frac{7}{27}$
Vậy: $0\leq xy+yz+zx-2xyz\leq \frac{7}{27}$

Bài 100915

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Bài 100915

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