Let $a,b,c>0$ such $ \frac{b+c}{a}+ \frac{c+a}{b}+ \frac{a+b}{c} = 2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} \right)$

Question

Let $a$, $b$ and $c$ be positive numbers such that $$ \dfrac{b+c}{a}+ \dfrac{c+a}{b}+ \dfrac{a+b}{c} = 2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac} \right).$$

Prove that

$$a^2+b^2+c^2+3\ge 2(ab+bc+ac)$$

my idea: let $$\dfrac{b}{a}=x,\dfrac{c}{b}=y,\dfrac{a}{c}=z$$

then I found this can't works

Answer 1

$\dfrac{b+c}{a} + \dfrac{c+a}{b} + \dfrac{a+b}{c} = 2\left(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}\right) \iff bc(b+c) + ac(a+c) + ab(a+b) = 2(a+b+c) \iff bc(b+c+a) - abc + ac(a+c+b) - abc + ab(a+b+c) - abc = 2(a+b+c) \iff (a+b+c)(ab+bc+ca) - 3abc = 2(a+b+c) \iff (a+b+c)(ab+bc+ca - 2) = 3abc$

Apply AM-GM inequality we have:

$(a+b+c)(ab+bc+ca - 2) = 3abc \leq \dfrac{(a+b+c)^3}{9}$, and divide both sides by $a+b+c > 0$ we get:

$ab+bc+ca - 2 \leq \dfrac{(a+b+c)^2}{9}$. This yields:

$18 \geq 9(ab+bc+ca) - (a+b+c)^2$, and divide by $6$ both sides to get:

$3 \geq \dfrac{3}{2}(ab+bc+ca) - \dfrac{(a+b+c)^2}{6}$. Using this we have:

$a^2 + b^2 + c^2 + 3 \geq a^2 + b^2 + c^2 + \dfrac{3}{2}(ab+bc+ca) - \dfrac{(a+b+c)^2}{6} \geq 2(ab+bc+ca) \iff a^2 + b^2 + c^2 - ab - bc - ca \geq 0 \iff \dfrac{(a-b)^2 + (b-c)^2 + (c-a)^2}{2} \geq 0$, which is clearly true.

Answer 2

The condition gives $\frac{\sum\limits_{cyc}(a^2b+a^2c)}{2(a+b+c)}=1$.

Thus, we need to prove that $$a^2+b^2+c^2+\frac{3\sum\limits_{cyc}(a^2b+a^2c)}{2(a+b+c)}\geq2(ab+ac+bc)$$ or $$\sum_{cyc}(2a^3+a^2b+a^2c-4abc)\geq0,$$ which is obviously true.

Done!