(5)

Ukraine 2008:
Cho các số thực dương $a,b,c$ thỏa: $a^2+b^2+c^2=3$. C/m: 
   $\sqrt{\frac{a^2}{a^2+b+c}}+\sqrt{\frac{b^2}{b^2+a+c}}+\sqrt{\frac{c^2}{c^2+a+b}}\leq \sqrt{3}$
 ad BĐT C-S
ta có $(a^{2}+b+c)(1+b+c)\geq (a+b+c)^{2}$
$\Rightarrow \sqrt{\frac{a^{2}}{a^{2}+b+c}}\leq \frac{a\sqrt{b+c+1}}{a+b+c}=\frac{a\sqrt{3(b+c+1)}}{(a+b+c)\sqrt{3}}\leq \frac{a+(b+c+1+3)}{2(a+b+c)\sqrt{3}}$
TT $\Rightarrow VT\leq \frac{4(a+b+c)+2(ab+bc+ca)}{2(a+b+c)\sqrt{3}}=\frac{2\sqrt{3}}{3}+\frac{1}{\sqrt{3}} \frac{ab+bc+ca}{a+b+c}=\frac{2\sqrt{3}}{3}+\frac{1}{3}\sqrt{ab+bc+ca}\leq \sqrt{3}$
Dấu "=" $\Leftrightarrow a=b=c=1$
uk.chết đợi tí tui sửa –  Nguyễn Nhung 24-08-16 08:58 PM
từ dòng thứ 2 ý –  tran85295 24-08-16 08:55 PM
thì ad nó <= á –  Nguyễn Nhung 24-08-16 08:50 PM
ở đây cho $\sum a^2=3$ mà, phải $\sum a=3$ đâu :)) –  tran85295 24-08-16 06:59 PM
Sử dụng BĐT cauchy schwarz
(a2+b+c)(1+b+c)(a+b+c)2
áp dụng cauchy schwarz 1 lần nữa ta dc
a1+b+c+b1+c+a+c1+a+ba+b+c3
VT=aa(1+b+c)+bb(1+c+a)+cc(1+a+b)a+b+c(a+b+c)(a(1+b+c)+b(1+c+a)+c(1+a+b))a+b+c=1+2(ab+bc+ca)a+b+c1+2(a+b+c)33
vậy ta có đpcm 
nguồn diendantoanhoc.net –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 24-08-16 05:33 PM

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