Cho $a;b;c>0$
Chứng minh: $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}$
Cho $a;b;c>0$
Chứng minh: $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}$
Cho $a;b;c>0$
Chứng minh: $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}$
Ta có $\sum \frac{a^2}{b}=\sum (\frac{a^2}{b}-a+b)=\sum \frac{a^2-ab+b^2}{b}\geq \frac{(\sum \sqrt{a^2-ab+b^2})^2}{\sum a}\geq \sum \sqrt{a^2-ab+b^2}< = > \sum \sqrt{a^2-ab+b^2}\geq \sum a$
BDT đúng vì $\sum \sqrt{a^2-ab+b^2}=\sum \sqrt{\frac{3}{4}(a-b)^2+\frac{1}{4}(a+b)^2}\geq \sum \sqrt{\frac{1}{4}(a+b)^2}=\sum \frac{1}{2}(a+b)=\sum a$
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