Bài toán:
Cho $$\left\{\begin{matrix} 0< a\leq b\leq c & \\ 0<
x,y,z& \end{matrix}\right.$$
Cmr: $$(xa+yb+zc)\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)\leq
\dfrac{(a+c)^{2}}{4ac}(x+y+z)^{2}$$
$f(x)=x^2-(a+c)x+ac=0$ 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)
(đpcm)
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