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C

chonhoi110

Bđt cần chứng minh $\leftrightarrow \dfrac{3}{xy+yz+zx}+\dfrac{2}{x^2 + y^2 + z^2}-14 >0$

$\leftrightarrow \dfrac{3}{xy+yz+zx}+\dfrac{2}{(x + y + z)^2-2(xy+yz+zx)} - 14 >0$

$\leftrightarrow \dfrac{3}{xy+yz+zx}+\dfrac{2}{1-2(xy+yz+zx)} - 14>0$ (vì $x+y+z =1$) (*)

Đặt $a=xy+yz+zx$

(*) $\leftrightarrow \dfrac{3}{a}+\dfrac{2}{1-2a} - 14 >0$

$\leftrightarrow 3(1-2a)+2a-14a(1-2a) >0$
$\leftrightarrow 28a^2-18a+3 >0 \rightarrow$ Đpcm
 
R

ronaldover7

x,y,z >= 0 nha bạn,chứ ko thì x=-100,y=100,z=0 sao >14 được

Ta có : x,y,z \geq 0 \Rightarrow xy+yz+zx \geq 0
Cm được $x^2$+$y^2$+$z^2$ \geq xy+yz+zx
Ta có:$(x+y+z)^2$=$x^2$+$y^2$+$z^2$+2(xy+yz+zx)=1 \geq 3(xy+yz+zx)
\Rightarrow $\frac{1}{3}$ \geq xy+yz+zx
Cm được:$\frac{1}{a}$+$\frac{1}{b}$ \geq $\frac{4}{a+b}$
Đặt xy+yz+zx=a,$x^2$+$y^2$+$z^2$=b
Ta có:$\frac{3}{a}$+$\frac{2}{b}$
=$\frac{2}{a}$+2($\frac{1}{2a}$+$\frac{1}{b}$)
=$\frac{2}{a}$+2.$\frac{4}{b+2a}$=$\frac{2}{a}$+8
Lại có $\frac{1}{3}$ \geq xy+yz+zx
\Rightarrow $\frac{1}{ xy+yz+zx }$ \geq 3
\Rightarrow $\frac{2}{ xy+yz+zx }$ \geq 6
\Rightarrow $\frac{2}{a}$ \geq 6
\Rightarrow $\frac{2}{a}$+8 \geq 14
\Rightarrow dpcm​
 
C

congchuaanhsang

Cauchy - Schwarz:

$\dfrac{3}{xy+yz+xz}+\dfrac{2}{x^2+y^2+z^2}=\dfrac{6}{2xy+2yz+2xz}+\dfrac{2}{x^2+y^2+z^2}$ \geq $\dfrac{(\sqrt{6}+\sqrt{2})^2}{(x+y+z)^2}=(\sqrt{6}+\sqrt{2})^2$ > 14
 
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