$(xa+yb+zc)\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)\leq \dfrac{(a+c)^{2}}{4ac}(x+y+z)^{2}$

Bài toán:

Cho $$\left\{\begin{matrix} 0< a\leq b\leq c & \\ 0< x,y,z& \end{matrix}\right.$$

Cmr:  $$(xa+yb+zc)\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)\leq \dfrac{(a+c)^{2}}{4ac}(x+y+z)^{2}$$

Lời giải:

$f(x)=x^2-(a+c)x+ac=0$ có $2$ nghiệm là $a$ và $c$

Do $a\leq b\leq c$ nên $f(b)\leq 0$

$$\Leftrightarrow b^2-(a+c)b+ac\leq 0$$

$$\Leftrightarrow yb+\dfrac{acy}{b}\leq y(a+c)$$

$$\Rightarrow \left(xa+ac.\dfrac{x}{a}\right)+\left(yb+ac.\dfrac{y}{b}\right)+\left(zc+ac.\dfrac{z}{c}\right)\leq (a+c)(x+y+z)$$

$$\Rightarrow xa+yb+zc+ac.\sum \dfrac{x}{a}\leq (a+c)(x+y+z)~~~(1)$$

Áp dụng AM-GM có:

$$VT(1)\geq 2\sqrt{(xa+yb+zc).ac.\sum \dfrac{x}{a}}$$

$$\Rightarrow 2\sqrt{(xa+yb+zc).ac.\sum \dfrac{x}{a}}\leq (a+c)(x+y+z)$$

$$\Leftrightarrow 4(xa+yb+zc).ac.\sum \dfrac{x}{a}\leq (a+c)^2(x+y+z)^2$$

$$\Leftrightarrow (xa+yb+zc).\sum \dfrac{x}{a}\leq \dfrac{(a+c)^2}{4ac}(x+y+z)^2$$
(đpcm)