Does the following statement/inequality holds for $a,b,c \in \mathbb{R^+}$?
$$\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$$
I've been thinking for hours and I couldn't find a solution to this inequality. I tried to use Cauchy-Schwarz Inequality for fraction:
$$\frac{x^2}{a} + \frac{y^2}{b} + \frac{z^2}{c} \ge \frac{(z+x+y)^2}{a+b+c}$$
But I only get:
$$LHS \ge \frac{(\sqrt{a} + \sqrt{b} + \sqrt{c})^2}{(n+k)(a+b+c)}$$
And then I can't prove that:
$$(\sqrt{a} + \sqrt{b} + \sqrt{c})^2 \ge 3(a+b+c)$$ $$a + b + c + 2\sqrt{ab} + 2\sqrt{ac} + 2\sqrt{bc} \ge 3(a+b+c)$$ $$\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \ge a+b+c$$
Which simplifyies to:
$$\sqrt{a}(\sqrt{b} - \sqrt{a}) + \sqrt{b}(\sqrt{c} - \sqrt{b}) + \sqrt{c}(\sqrt{a} - \sqrt{c}) \ge 0$$
But if $a\neq b \neq c$, one of the terms must be negative, so nothing here.
And at last I plug it into Wolfram Alpha, and for some random numbers $n$ and $k$ it's true. I wonder maybe the fact that the LHS is cyclic we make the LHS to have minimal values when $a=b=c$?
By applying CBS is obtained
$$\frac{a^2}{na^2+kab} + \frac{b^2}{nb^2+kbc} + \frac{c^2}{nc^2+kca} \ge \frac{(a+b+c)^2}{n(a^2+b^2+c^2)+k(ab+bc+ca)}.$$
But
$$\frac{(a+b+c)^2}{n(a^2+b^2+c^2)+k(ab+bc+ca)}\ge \frac{3}{n+k}$$
is equivalent to
$$(k-2n)(a^2+b^2+c^2-ab-bc-ca)\ge 0$$
which is true under the additional condition
$$k\ge2n.$$