\(\left\{{}\begin{matrix}\left|x\right|\ge3\\\left|y\right|\ge3\\\left|z\right|\ge3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|\dfrac{1}{x}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{y}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}\end{matrix}\right.\)

\(\left|A\right|=\left|\dfrac{xy+yz+xz}{xyz}\right|=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\le\left|\dfrac{1}{x}\right|+\left|\dfrac{1}{y}\right|+\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)

\(\Rightarrow A\le\left|A\right|\le1\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=3\)

Đề là \(\frac{xy+yz+xz}{xyz}\le1\)  nhé!

Giải:

Ta có:

\(\left|H\right|=\left|\frac{xy+yz+xz}{xyz}\right|\le\frac{\left|xy\right|+\left|yz\right|+\left|xz\right|}{\left|xyz\right|}\)

\(=\frac{1}{\left|x\right|}+\frac{1}{\left|y\right|}+\frac{1}{\left|z\right|}\le\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)

Vậy \(H=\frac{xy+yz+xz}{xyz}\le1\) (Đpcm)

ta có :

\(\frac{ax+by}{2}\ge\frac{a+b}{2}.\frac{x+y}{2}\Leftrightarrow2\left(ax+by\right)\ge\left(a+b\right)\left(x+y\right)\)

\(\Leftrightarrow2\left(ax+by\right)\ge ax+ay+bx+by\)

\(\Leftrightarrow ax-ay+by-bx\ge0\Leftrightarrow\left(a-b\right)\left(x-y\right)\ge0\)

Điều này đúng do giả thuyết \(a\ge b,x\ge y\)

ta có \(\dfrac{ax+by}{2}\) ≥ \(\dfrac{a+b}{2}\). \(\dfrac{x+y}{2}\)

<=> 2(ax + by) ≥ (a + b)(x + y)

<=> 2(ax + by) ≥ ax + ay + bx + by

<=> ax + by - ay - bx ≥ 0

<=> (a - b)(x - y) ≥ 0 (luôn đúng vì giả thiết a ≥ b và x ≥ y)

vậy nếu a ≥ b, x ≥ y thì \(\dfrac{ax+by}{2}\) ≥ \(\dfrac{a+b}{2}\). \(\dfrac{x+y}{2}\)

Ta có \dfrac{ax + by}2 \ge \dfrac{a + b}2 . \dfrac{x + y}22ax+by​≥ 2a+b​. 2x+ y​

$\iff 2(ax+by)\ge (a+b)(x+y)$

$\iff 2(ax+by)\ge ax+ay+bx+by$

$\iff ax+by-ay-bx\ge 0$

$\iff (a-b)(x-y)\ge 0$ (luôn đúng vì giả thiết $a\ge b$ và $x\ge y$).

Vậy nếu $a\ge b$, $x\ge y$ thì \dfrac{ax + by}2 \ge \dfrac{a + b}2 . \dfrac{x + y}22ax+by​≥ 2a+b​. 2x+ y​.

\(x+y=1\)

Áp dụng BĐT AM-GM, ta có:

\(\dfrac{x^2}{1}+\dfrac{y^2}{1}\ge\dfrac{\left(x+y\right)^2}{2}=\dfrac{1^2}{2}=\dfrac{1}{2}\)

--> \(x^2+y^2\ge\dfrac{1}{2}\)

\(P=a^3+b^3=\left(a+b\right)\left(\left(a+b\right)^2-3ab\right)\ge\left(a+b\right)\left[\left(a+b\right)^2-\frac{3}{4}\left(a+b\right)^2\right]\)

\(P\ge\frac{1}{4}\left(a+b\right)\left(a+b\right)^2=\frac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)

-Đề sai.

Đề này sai đó bạn.

Giả sử c = 2,5; a = 2 và c = 1,5

Ta có: \(c\ge a;c\ge b\) nhưng \(c< a+b\) (mâu thuẫn với đề bài).

\(\ge\)0 nhá

Ta có: \(x-y+z=0\)
    \(\Rightarrow\left(x-y+z\right)^2=0 \)
  \(\Rightarrow\left(x-y+z\right).\left(x-y+z\right)=0\)
   \(\Rightarrow x\left(x-y+z\right)-y\left(x-y+z\right)+z\left(x-y+z\right)=0\)
   \(\Rightarrow x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=0\)
  \(\Rightarrow x^2+y^2+z^2=xy+xy+yz+yz-xz-xz\)
   \(\Rightarrow x^2+y^2+z^2=2xy+2yz-2xz\)
   \(\Rightarrow x^2+y^2-z^2=2\left(xy+yz-xz\right)\)
Mà: \(x^2+y^2-z^2\ge0\)
\(\Rightarrow2\left(xy+yz-xz\right)\ge0\)
\(\Rightarrow xy+yz-xz\ge0\)(đpcm)
   Vậy: \(xy+yz-xz\ge0\)