Minimize The Expression 7

Use substitution $(x,y,z) = (2\cos A , 2\cos B, 2\cos C)$ . It is well known that $x^2 + y^2 + z^2 + xyz = 4$ .

The expression becomes $(\frac{x}{y})^2 + (\frac{y}{z})^2 + (\frac{z}{x})^2 + xyz$

---

$\textbf{(Lemma 1)}$ If $abc \leq 1$ , $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b + c$

$\textit{Proof : }$ Multiply $abc$ $a^2c + b^2a + c^2b \geq abc(a+b+c)$ We prove a stronger inequality $a^2c + b^2a + c^2b \geq (abc)^{\frac{2}{3}}(a+b+c)$ $a^2c + b^2a + c^2b \geq a^{\frac{5}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}} + a^{\frac{2}{3}}b^{\frac{5}{3}}c^{\frac{2}{3}} + a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{5}{3}}$ By AM-GM $a^2c + a^2c + b^2a \geq 3a^{\frac{5}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}}$ $b^2a + b^2a + c^2b \geq 3a^{\frac{2}{3}}b^{\frac{5}{3}}c^{\frac{2}{3}}$ $c^2b + c^2b + a^2c \geq 3a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{5}{3}}$ Adding all of these yields the desired Lemma.

---

$\textbf{(Lemma 2)}$ $\cos A \cos B \cos C \leq \frac{1}{8}$ $\textit{Proof:}$ Take $log$ of both sides (natural logarithm). Suffices to prove $\log \cos A + \log \cos B + \log \cos C \leq \log \frac{1}{8}$ Consider the function $f(x) = \log \cos x$

$f'(x) = -\tan x$

$f''(x) = -\sec^2 x \implies$ concave for $[0, \frac{\pi}{2} ]$ .

By Jensen, $f(A) + f(B) + f(C) \leq 3f(\frac{A+B+C}{3}) = 3f(60^{\circ}) = \log \frac{1}{8}$ and we have proven the Lemma.

---

$(\frac{x}{y})^2 + (\frac{y}{z})^2 + (\frac{z}{x})^2 + xyz$ Let $(a,b,c) = (x^2 , y^2, z^2)$ and notice that $abc \leq 1$ since it all goes down to $\cos A \cos B \cos C \leq \frac{1}{8}$

This implies that $(\frac{x}{y})^2 + (\frac{y}{z})^2 + (\frac{z}{x})^2 + xyz \geq x^2 + y^2 + z^2 + xyz = \boxed{4}$

from our first identity. Equality holds when $A = B = C = 60^{\circ}$