Chủ đề này có 2 trả lời

[tritanngo99]

Cho $x,y,z>0$ thỏa mãn: $x+y+z=3$. CMR: $P=(x+2y+3z)(x+\frac{y}{2}+\frac{z}{3})\le 12$.

Tổng quát cho dạng này:

Cho $0<a\le b\le c$ và $0<x,y,z$. Chứng minh rằng:

$(ax+by+cz)(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})\le \frac{(a+c)^2}{4ac}(x+y+z)^2$

[phamhuy1801]

$(x+2y+3z)(x+ \frac{y}{2} + \frac{z}{3}) = \frac{1}{3}.(x+2y+3z)(3x+ \frac{3y}{2} + z) \le \frac{1}{3}.\frac{(x+2y+3z+3x+1,5y+z)^2}{4} = \frac{1}{12}.(4x+3,5y+4z)^2 < \frac{1}{12}.(4x+4y+4z)^2=12$

[tritanngo99]

$(x+2y+3z)(x+ \frac{y}{2} + \frac{z}{3}) = \frac{1}{3}.(x+2y+3z)(3x+ \frac{3y}{2} + z) \le \frac{1}{3}.\frac{(x+2y+3z+3x+1,5y+z)^2}{4} = \frac{1}{12}.(4x+3,5y+4z)^2 < \frac{1}{12}.(4x+4y+4z)^2=12$

Mình chứng minh trường hợp tổng quát như sau:

Đặt $f(x)=x^2-(a+c)x+ac=0$ có 2 nghiệm $a,c$.

Mà $a\le b\le c\implies f(b)\le 0\iff b^2-(a+c)b+ac\le 0$

$\iff b+\frac{ac}{b}\le a+c\iff yb+ac\frac{y}{b}\le (a+c)y$.

$\iff (xa+ac\frac{x}{a})+(yb+ac\frac{y}{b})+(zc+ca\frac{z}{c})\le (a+c)x+(a+c)y+(a+c)z$

$\implies xa+yb+zc+ac(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})\le (a+c)(x+y+z)(1)$.

Theo BDT CauChy ta có:

$2\sqrt{(xa+yb+zc)ac(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})}\le (a+c)(x+y+z)(\text{ do (1)})$

$\iff 4(xa+yb+zc)ac(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})\le (a+c)^2(x+y+z)^2$

$\iff (xa+yb+zc)(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})\le \frac{(a+c)^2}{4ac}(x+y+z)^2(dpcm)$