Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\sum \frac{c^{2013}}{a+b-c}=\sum \frac{c^{4024}}{ac^{2011}+bc^{2011}-c^{2012}}\geq \frac{(\sum a^{2012})^2}{a^{2011}(b+c)+b^{2011}(c+a)+c^{2011}(b+a)-\sum a^{2012}}\)
Ta sẽ CM:
\(a^{2011}(b+c)+b^{2011}(c+a)+c^{2011}(b+a)-\sum a^{2012}\leq \sum a^{2012}\)
\(\Leftrightarrow a^{2011}(a-b)+a^{2011}(a-c)+b^{2011}(b-a)+b^{2011}(b-c)+c^{2011}(c-a)+c^{2011}(c-b)\geq 0\)
\(\Leftrightarrow \sum (a-b)(a^{2011}-b^{2011})\geq 0\Leftrightarrow \sum (a-b)^2(a^{2010}+...+b^{2010})\geq 0\) (luôn đúng)
Do đó: \(\sum \frac{c^{2013}}{a+b-c}\geq \frac{(\sum a^{2012})^2}{\sum a^{2012}}=\sum a^{2012}\)
Dấu "=" xảy ra khi $a=b=c$. Tức là $ABC$ là tam giác đều.
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