$P=\dfrac{ab+bc+ca}{a^2+b^2+c^2}+\dfrac{(a+b+c)^3}{abc}$

Bài toán:
Cho $a;b;c>0$. Tìm $Min$:
$$P=\dfrac{ab+bc+ca}{a^2+b^2+c^2}+\dfrac{(a+b+c)^3}{abc}$$



Lời giải:


$$P=\dfrac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}+\dfrac{(a+b+c)(a^{2}+b^{2}+c^{2})}{9abc}+\dfrac{8(a+b+c)(a^{2}+b^{2}+c^{2})}{9abc}+\dfrac{2(ab+bc+ca)(a+b+c)}{abc}$$

Áp dụng BĐT Cô-si cho 3 số dương ta có:
$$(a+b+c)(a^{2}+b^{2}+c^{2})\geq 3\sqrt[3]{abc}.3\sqrt[3]{a^2b^2c^2}=9abc$$

$$(a+b+c)(ab+bc+ca)\geq 3\sqrt[3]{abc}.3\sqrt[3]{a^2b^2c^2}=9abc$$

$$\Rightarrow P\geq 2\sqrt{\dfrac{(a+b+c)(ab+bc+ca)}{9abc}}+\dfrac{8.9abc}{9abc}+\dfrac{2.9abc}{abc}$$
$$\geq 2+8+18=28$$