Given a²+b²+c² = 1 show me that $\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2 > 4$

Question

Given $a^2+b^2+c^2 = 1$ and $a, b$ and $c$ are real positive numbers, shown me that

$$\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2 > 4$$

Answer 1

Without loss of generality assume $0<a\leq b\leq c$. Then as $b+c\geq 2a$, we have $(\frac{b+c}{a})^2\geq 4$ and as rest of two squares are positive we have the inequality $$\left(\frac{a+b}{c}\right)^2+\left(\frac{b+c}{a}\right)^2+\left(\frac{c+a}{b}\right)^2>4$$.

Answer 2

First consider (prove) that $x + \frac{1}{x} \ge 2$ for $x \gt 0$.

Then expand the squares and note that you can regroup 3 terms of the form $\frac{a^2}{b^2} + \frac{b^2}{a^2}$. Each of those will then be $\ge 2$ so the sum will be $\ge 6$ which implies $\gt 4$..

Answer 3

\begin{align*} \left(\frac{a+b}{c} \right)^2 + \left(\frac{b+c}{a} \right)^2 +\left(\frac{c+a}{b} \right)^2 &\ge \frac{(a+b+b+c+c+a)^2}{a^2+b^2+c^2}\\ &=\frac{4(a+b+c)^2}{a^2+b^2+c^2} \ge 4. \end{align*}