Bài 101518

Question

Cho $1\leq x\leq y\leq z\leq 2$.Chứng minh rằng:
$1/y\left ( \frac{1}{x} + \frac{1}z\right )+ \frac{1}{y}\left ( x +z \right )\leq \left ( \frac{1}{x} + \frac{1}z\right )\left ( x +z \right )$
$2/\left ( x +y +z \right )\left ( \frac{1}{x}+ \frac{1}{y}+ \frac{1}z\right )\leq 10$

score 0 · Answer 1

$1/$BĐT: $\Leftrightarrow y\frac{x+z}{xz}+\frac{x+z}{y}\leq \frac{\left ( x +z \right )^{2}}{xz}$
$\Leftrightarrow y^{2}+xz\leq y\left ( x +z \right )$ ( vì $x, y, z>0$)
$\Leftrightarrow y^{2}-\left ( x +z \right )y+xz\leq 0$
$\Leftrightarrow \left ( y-x\right ) \left ( y-z\right )\leq 0$, vì $\begin{cases}y-x\geq 0 \\ y-z\leq 0\end{cases}$
Nên BĐT cuối luôn đúng $\Rightarrow $ (ĐPCM)
$2/$Ta có:
$\begin{cases}\left ( 1-\frac{x}{y} \right )\left ( 1-\frac{y}{z} \right )\geq 0\\ \left ( 1-\frac{y}{x} \right )\left ( 1-\frac{z}{y} \right )\geq 0 \end{cases}\Rightarrow \frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}\leq 2+\frac{x}{z}+\frac{z}{x}$
$\Rightarrow $ VT $= 3+ \frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}$
$\leq 5+2\left ( \frac{x}{z}+\frac{z}{x}\right )\leq 10+\left ( \frac{x}{z}-2\right )\left ( 2-\frac{z}{x}\right )\leq 10$ (vì $\frac{x}{z}, \frac{z}{x} \leq 2)$
$\Rightarrow $ ĐPCM

Bài 101518

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Bài 101518

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