Cho $x,y,z$ là $3$ số thực dương. Chứng minh:
$\frac{x}{y+z}+\sqrt{\frac{x}{y+z}}+\frac{y}{x+z}+\sqrt{\frac{y}{x+z}}+\frac{z}{x+y}+\sqrt{\frac{z}{x+y}}>3$
Theo bđt $Nesbit: \frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y} \geq \frac{3}{2} (*)$ 
thật vậy 
$VT+3=\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1=(x+y+z)(\frac1{x+y}+\frac1{y+z}+\frac1{z+x})$
Nếu đặt $x+y=a;y+z=b;z+x=c$ thì có $VT+3=\frac{a+b+c}2(\frac1a+\frac1b+\frac1c)\geq \frac32\sqrt[3]{abc}.3\sqrt[3]{\frac1{abc}}=\frac92hay VT\ge\frac32$
Cách khác, BĐT này có thể bằng CM theo BĐT BCS dạng Engel
$VT=\frac{x^2}{xy+xz}+\frac{y^2}{yz+yx}+\frac{z^2}{zx+zy}\ge\frac{(x+y+z)^2}{2(xy+yz+zx)}$
Áp dụng BĐT quen thuộc sau $(x+y+z)^2\ge3(xy+yz+zx)\Leftrightarrow(a-b)^2+(b-c)^2+(c-a)^2\ge0$
thì ta đã CM đc BĐT Nesbit
Ta lại có:
 $x+(y+z) \overset{AM-GM}{\geq} 2\sqrt{x(y+z)}$
$ \Leftrightarrow \sqrt{x}(x+y+z) \geq 2x\sqrt{y+z}$
$\Leftrightarrow \sqrt{\frac{x}{y+z}} \geq \frac{2x}{x+y+z}$
Tương tự $\Leftrightarrow \sqrt{\frac{y}{x+z}} \geq \frac{2y}{x+y+z}$;$\Leftrightarrow \sqrt{\frac{z}{x+y}} \geq \frac{2x}{x+y+z}$
Cộng 3 Bđt $\Rightarrow \sqrt{\frac{x}{y+z}} +  \sqrt{\frac{y}{x+z}}+ \sqrt{\frac{z}{x+y}}>\frac{2(x+y+z)}{x+y+z}=2(**)$ (ở đây đẳng thức ko thể xảy ra vì 3 bđt trên ko đồng thời xảy ra dấu =)
Từ $(*),(**)\Rightarrow VT >\frac{3}{2}+2>3$ (đpcm)

ok bạn :) –  Moon 13-11-15 09:33 PM
vào kênh chat đi nói cho dễ –  tran85295 13-11-15 09:31 PM
ms học cosi và bunhia thôi bạn –  Moon 13-11-15 09:30 PM
bạn biết bđt (a cộng b cộng c)(1/a cộng 1/b cộng 1/c) lớn hơn hoặc = 9 ko –  tran85295 13-11-15 09:21 PM
ok thế để mình c/m cho –  tran85295 13-11-15 09:18 PM
mk chưa hk bđt nesbit –  Moon 13-11-15 09:16 PM

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