Lời giải:
Ta có:
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}=\frac{a^2}{b}-a+b+\frac{b^2}{c}-b+c+\frac{c^2}{a}-c+a\)
\(=\frac{a^2-ab+b^2}{b}+\frac{b^2-bc+c^2}{c}+\frac{c^2-ac+a^2}{a}\)
Áp dụng BĐT AM-GM:
\(\frac{a^2-ab+b^2}{b}+b\geq 2\sqrt{a^2-ab+b^2}\)
\(\frac{b^2-bc+c^2}{c}+c\geq 2\sqrt{b^2-bc+c^2}\)
\(\frac{c^2-ca+a^2}{a}+a\geq 2\sqrt{c^2-ca+a^2}\)
Cộng theo vế:
\(\Rightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+(a+b+c)\geq 2(\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2})\)
\(\Rightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq 2(\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2})-(a+b+c)(*)\)
Mà theo BĐT AM-GM:
\(\sqrt{a^2-ab+b^2}=\sqrt{(a+b)^2-3ab}\geq \sqrt{(a+b)^2-3.\frac{(a+b)^2}{4}}=\frac{a+b}{2}\)
Tương tự:....
\(\Rightarrow \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\geq \frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}=a+b+c(**)\)
Từ \((*); (**)\Rightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\)
Ta có đpcm. Dấu "=" xảy ra khi $a=b=c$
