Cho $a,b,c>0$ và $a+b+c=3$. CMR:
1) $\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\geqslant 1$ 
2) $\frac{a+1}{b^2+1}\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\geqslant 3$ 
10 ban ak –  lazuzu123 01-04-13 05:40 PM
toan nay la cua lop may vay ban? –  tamnguyen140698 30-03-13 11:11 AM
1.
Áp dụng BĐT Cauchy ta có:
    $\dfrac{a^2}{a+2b^2}$
$=a-\dfrac{2ab^2}{a+b^2+b^2}$
$\ge a-\dfrac{2ab^2}{3\sqrt[3]{ab^4}}$
$=a-\dfrac{2}{3}\sqrt[3]{a^2b^2}$
$\ge a-\dfrac{2}{9}(a+b+ab)$
Tương tự: $\dfrac{b^2}{b+2c^2}\ge b-\dfrac{2}{9}(b+c+bc);\dfrac{c^2}{c+2a^2}\ge c-\dfrac{2}{9}(c+a+ca)$
Cộng các BĐT trên lại ta được:
    $\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}$
$\ge a+b+c-\dfrac{2}{9}(a+b+ab+b+c+bc+c+a+ac)$
$=\dfrac{5}{9}(a+b+c)-\dfrac{2}{9}(ab+bc+ca)$
$\ge\dfrac{5}{9}(a+b+c)-\dfrac{2}{27}(a+b+c)^2=1$
Dấu bằng xảy ra khi: $a=b=c=1$
Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! –  khangnguyenthanh 17-02-13 05:04 PM

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