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[gagaga]

cho a,b,c>0. CMR:

$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq \sqrt{a^{2}-ab+b^{2}}+\sqrt{b^{2}-bc+c^{2}}+\sqrt{c^{2}-ca+a^{2}}$

[Khoa Linh]

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

Hoàn toàn tương tự rồi cộng vào ta có:

$VT\geq 2\left ( \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2} \right )-(a+b+c) $

Mặt khác 

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

$\Rightarrow 2\left ( \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2} \right )-(a+b+c)\geq \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}$

Edited by Khoa Linh, 17-02-2018 - 00:44.