[Toán 10]BĐT giúp mình làm bài này

H

huytrandinh

[TEX]T=\sum \frac{x^{2}}{xy+xz}[/TEX]
[TEX]\geq \frac{(x+y+z+t)^{2}}{xy+xz+yz+yt+zt+zx+tx+ty} (BCS)[/TEX]
[TEX]=\frac{(x+y+z+t)^{2}}{(y+z)(x+t)+(z+t)(x+y)}[/TEX]
[TEX].(y+z)(x+t)\leq \frac{(x+y+z+t)^{2}}{4}[/TEX]
[TEX].(z+t)(x+y)\leq \frac{(x+y+z+t)^{2}}{4}[/TEX]
[TEX]=>T\geq 2<=>x=y=z=t[/TEX]
 
N

noinhobinhyen

Xét :

$M=\dfrac{y}{y+z}+\dfrac{z}{z+t}+\dfrac{t}{t+x}+ \dfrac{x}{x+y}$

$N=\dfrac{z}{y+z}+\dfrac{t}{z+t}+\dfrac{x}{t+x}+ \dfrac{y}{x+y}$

Ta có $M+N=\dfrac{y+z}{y+z}+\dfrac{z+t}{z+t}+\dfrac{t+x}{t+x}+\dfrac{x+y}{x+y}=4$

Gọi Vế trái là S

theo bđt Cô-Si ta có :

$M+S \geq 4 ; N+S \geq 4 \Rightarrow M+N+2S \geq 8 \Leftrightarrow 2S \geq 4 \Leftrightarrow S \geq 2$
 
Top Bottom