[tex]a^{5}+\frac{1}{a}\geq 2\sqrt{a^{5}.\frac{1}{a}}=2a^{2}\\ b^{5}+\frac{1}{b}\geq 2\sqrt{b^{5}.\frac{1}{b}}=2b^{2}\\ c^{5}+\frac{1}{c}\geq 2\sqrt{c^{5}.\frac{1}{c}}=2c^{2}\\ \Rightarrow a^{5}+\frac{1}{a}+b^{5}+\frac{1}{b}+c^{5}+\frac{1}{c}\geq 2(a^{2}+b^{2}+c^{2})\\ a^{2}+b^{2}\geq 2ab;b^{2}+c^{2}\geq 2bc;c^{2}+a^{2}\geq 2ca\\ \Rightarrow 2(a^{2}+b^{2}+c^{2})\geq 2(ab+bc+ca)\\ a+b+c=3\Rightarrow (a+b+c)^{2}\geq 9\\\Leftrightarrow a^{2}+b^{2}+c^{2}+2(ab+bc+ca)\geq 9\\ \Leftrightarrow 3(a^{2}+b^{2}+c^{2})\geq 9\\ \Leftrightarrow a^{2}+b^{2}+c^{2}\geq 3\\ \Rightarrow a^{5}+\frac{1}{a}+b^{5}+\frac{1}{b}+c^{5}+\frac{1}{c}\geq 2(a^{2}+b^{2}+c^{2})\geq 2.3=6[/tex]
Vậy [tex]a^{5}+b^{5}+c^{5}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 6[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
