Trích:
Nguyên văn bởi hung.vx Cho $a,b,c$ là các số thực dương thoả $a^2+b^2+c^2\ge a+b+c$. Chứng minh rằng $$a^3+b^3+c^3+\frac{8}{(a+b)(b+c)(c+a)}\geq 4.$$ |
Ta có:
$$ \frac{\left ( a^{3}+b^{3}+c^{3} \right )^{3}}{\left ( a+b \right )(b+c)(c+a)} \geq \frac{27}{8}\left ( \frac{a^2+b^2+c^2}{a+b+c} \right )^6$$
Theo AM-GM:
$$\left ( a+b \right )\left ( b+c \right )\left ( c+a \right )\leq \frac{8.\left ( a+b+c \right )^{3}}{27}$$
$$\left ( a+b+c \right )\left ( a^{3}+b^{3}+c^{3} \right )\geq \left ( a.\sqrt{a^{3}}+b.\sqrt{b^{3}}+c.\sqrt{c^{3}} \right )^{2}= \left ( a^{2}+b^{2}+c^{2} \right )^{2}$$
$$\left ( a+b+c \right )^3\left ( a^3+b^3+c^3 \right )^3\geq (a^2+b^2+c^2)^6$$
$$\Rightarrow \frac{\left ( a+b+c \right )^{3}\left ( a^{2}+b^{2}+c^{2}\right )^{3}}{\left ( a+b \right )\left ( b+c \right )\left ( c+a \right )}\geq \frac{27\left ( a^{2}+b^{2}+c^{2} \right )^{6}}{8\left ( a+b+c \right )^{3}}$$
Do đó:
$$ \frac{\left ( a^{3}+b^{3}+c^{3} \right )^{3}}{\left ( a+b \right )(b+c)(c+a)} \geq \frac{27}{8}\left ( \frac{a^2+b^2+c^2}{a+b+c} \right )^6$$
Sử dụng AM-GM:
$$a^{3}+b^{3}+c^{3}+\frac{8}{\left ( a+b \right )\left ( b+c \right )\left ( c+a \right )}= \frac{a^{3}+b^{3}+c^{3}}{3}+\frac{a^{3}+b^{3}+c^{3 }}{3}+\frac{a^{3}+b^{3}+c^{3}}{3}+\frac{8}{\left ( a+b \right )\left ( b+c \right )\left ( c+a \right )}$$
$$\geq 4\sqrt[4]{\frac{8\left ( a^{3} +b^{3}+c^{3}\right )^{3}}{27\left ( a+b \right )\left ( b+c \right )\left ( c+a \right )}}\geq 4\sqrt[4]{\frac{8}{27}.\frac{27\left ( a^{2}+b^{2}+c^{2} \right )^{6}}{8\left ( a+b+c \right )^{6}}}\geq 4 $$
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]