Question

Cho a, b, c > 0

Cmr: \(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\)

Answer 1

Lời giải:

Áp dụng BĐT AM-GM:

\(\frac{a^2}{b}=\frac{a^2-ab+b^2}{b}+a-b=\frac{a^2-ab+b^2}{b}+b+(a-2b)\geq 2\sqrt{a^2-ab+b^2}+(a-2b)\)

Tương tự:

\(\frac{b^2}{c}\geq 2\sqrt{b^2-bc+c^2}+(b-2c)\)

\(\frac{c^2}{a}\geq 2\sqrt{c^2-ca+a^2}+(c-2a)\)

Cộng theo vế:
\(\sum \frac{a^2}{b}\geq 2\sum \sqrt{a^2-ab+b^2}-(a+b+c)(1)\)

Mà theo BĐT AM-GM:

\(\sqrt{a^2-ab+b^2}=\sqrt{(a+b)^2-3ab}\geq \sqrt{(a+b)^2-\frac{3}{4}(a+b)^2}=\frac{a+b}{2}\)

\(\Rightarrow \sum \sqrt{a^2-ab+b^2}\geq \sum \frac{a+b}{2}=a+b+c(2)\)

Từ $(1);(2)\Rightarrow \sum \frac{a^2}{b}\geq \sum \sqrt{a^2-ab+b^2}$ (đpcm)

Dấu "=" xảy ra khi $a=b=c$