$a)$
Áp dụng BĐT Cauchy ta có:
$\dfrac{a}{b}+\dfrac{a}{b}+\dfrac{b}{c}\ge 3\sqrt[3]{\dfrac{a^2}{bc}}=\dfrac{3a}{\sqrt[3]{abc}}$
Chứng minh tương tự:
$\dfrac{b}{c}+\dfrac{b}{c}+\dfrac{c}{a}\ge \dfrac{3b}{\sqrt[3]{abc}}$
$\dfrac{c}{a}+\dfrac{c}{a}+\dfrac{a}{b}\ge \dfrac{3c}{\sqrt[3]{abc}}$
Cộng vế theo vế
$\Rightarrow 3(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a})\ge 3.\dfrac{a+b+c}{\sqrt[3]{abc}}\\ \Rightarrow \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge \dfrac{a+b+c}{\sqrt[3]{abc}}$
Áp dụng vào bài toán, kết hợp BĐT Cauchy:
$\Rightarrow \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{3\sqrt[3]{abc}}{a+b+c}\ge \dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{3\sqrt[3]{abc}}{a+b+c}\\ \ge \dfrac{2}{3}.\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{a+b+c}{3\sqrt[3]{abc}}+\dfrac{3\sqrt[3]{abc}}{a+b+c}\\ \ge \dfrac{2}{3}.\dfrac{3\sqrt[3]{abc}}{3\sqrt[3]{abc}}+2\sqrt{1}=\dfrac{2}{3}.3+2=4(đpcm)$
Đẳng thức khi $a=b=c$
Câu $b)$ tương tự nhé
$c)$ Áp dụng BĐT phụ (tự cm):
$(a+b+c)^2\ge 3(ab+bc+ca)=3(a+b+c)$
$\Rightarrow a+b+c\ge 3$
Ta cần CM: $\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge \dfrac{(a+b+c)(a^2+b^2+c^2)}{ab+bc+ca}$
$\Leftrightarrow (ab+bc+ca)(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a})\ge (a+b+c)(a^2+b^2+c^2)\\ \Leftrightarrow \dfrac{a^3c}{b}+\dfrac{b^3a}{c}+\dfrac{c^3a}{b}\ge a^2b+b^2c+c^2a$
Theo Cauchy: $\dfrac{a^3}{b}+\dfrac{b^3a}{c}\ge 2\sqrt{a^4b^2}=2a^2b$
Tương tự, cộng lại và chia đôi 2 vế ta được đpcm
Áp dụng vào bài toán:
$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+\dfrac{3n}{a^2+b^2+c^2}\ge \dfrac{(a+b+c)(a^2+b^2+c^2)}{ab+bc+ca}+\dfrac{3n}{a^2+b^2+c^2}\\ \Leftrightarrow VT\ge a^2+b^2+c^2+\dfrac{3n}{a^2+b^2+c^2}(a+b+c=ab+bc+ca)\\ \Leftrightarrow VT\ge 3n(\dfrac{1}{a^2+b^2+c^2}+\dfrac{a^2+b^2+c^2}{9})+(\dfrac{3-n}{3})(a^2+b^2+c^2)\\ \Leftrightarrow VT\ge 3n.2\sqrt{\dfrac{1}{9}}+\dfrac{3-n}{n}.3=3+n(đpcm)$
Đẳng thức xảy ra khi $a=b=c=1$ và $n=3$
