Cho $3$ số dương $a,b,c$ thõa mãn đk:  $$a^2+b^2+c^2=3$$
C/m : $\color{red}{\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\geq3}$

Câu hỏi này được treo giải thưởng trị giá +1000 vỏ sò bởi thuhuong1607hhpt@gmail.com, đã hết hạn vào lúc 03-09-15 04:53 PM

Cách 3: 



ảo thía @@ :)) –  Confusion 09-06-16 02:00 PM
Cách 2:
Nhân 2 vế (của bất ở cách 1) với $(x+y+z).$
A/D Bunyakovsky, ta có:
$\frac{(a^2+b^2)(a+b+c)}{a+b}=a^2+b^2+\frac{c(a^2+b^2)}{a+b}\geq a^2+b^2+\frac{c.\frac{(a+b)^2}{2}}{a+b}=a^2+b^2+\frac{c(a+b)}{2}$
Tương tự:
......................................
........................................
Cộng lại được:
$VT\geq 2(a^2+b^2+c^2)+(ab+bc+ca)=$$\frac{3(a^2+b^2+c^2)+(a+b+c)^2}{2}$
Vậy cần c/m:
Hồng $\geq (a+b+c)\sqrt{3(a^2+b^2+c^2)}$
Cái này lđ theo Cauchy
 


Ta c/m: $\Sigma \frac{a^2+b^2}{a+b}\geq \sqrt{3(\Sigma a^2)}$

Cách 1:
$\Sigma (\frac{a^2+b^2}{a+b}-\frac{a+b}{2})\geq \sqrt{3(\Sigma a^2)}-(\Sigma a)$
$\Leftrightarrow \Sigma \frac{(a-b)^2}{2(a+b)}\geq \frac{\Sigma (a-b)^2}{\sqrt{3(\Sigma a^2)}+(\Sigma a)}$
Mà:
$(a+b+c)^2\leq 3(a^2+b^2+c^2)$
nên:
$\sqrt{3(a^2+b^2+c^2)}+a+b+c\geq 2(a+b+c)$
Ta cần c/m:
$\Sigma \frac{(a-b)^2}{2(a+b)}\geq \frac{\Sigma (a-b)^2}{2(a+b+c)}$
$\Leftrightarrow \Sigma \frac{c(a-b)^2}{a+b}\geq 0$ lđ!
$\Rightarrow ..............$
Đẳng thức khi $a=b=c./$
max r! mai lm tiếp ==" –  Confusion 04-06-16 06:37 PM
To be continue !! –  ☼SunShine❤️ 04-06-16 06:15 PM
pp SOS à –  tran85295 04-06-16 04:06 PM

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