Interesting Inequality using AM-GM and other identities. Let $a, b, c > 0$. Prove that $$\sqrt{a^2-ab+b^2} + \sqrt{b^2 - bc + c^2} + \sqrt{c^2 - ca + a^2} \le \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b}.$$ This should be solvable with AM-GM and a few other inequalities, but I am a little stuck on this problem.
My idea was to remove the radical. $\sqrt{a^2-ab+b^2} \le \frac{a^2-ab+b^2}{a+b} + \frac{a+b}{4}$ by AM-GM. Adding this up cyclically, it suffices to show the inequality
$$\frac{5}{2}\left(a+b+c\right)-3\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\right) \le \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b},$$
which I'm pretty sure is true, but I have no clue how to prove.
This inequality resembles https://artofproblemsolving.com/community/c6h1288310p6804993 and https://artofproblemsolving.com/community/q2h1817483p12130020, the latter of which is a weaker version of this inequality.

The inequality
$$\frac{5}{2}\left(a+b+c\right)-3\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\right) \le \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b},$$
doesn't true, example $a = \frac 13,\, b = \frac 1 4,\,c = 1.$
Let $a = \frac{1}{x},\,b = \frac{1}{y},\,c = \frac{1}{z},$ the inequality become
$$x \sqrt{y^{2}-y z+z^{2}}+y \sqrt{z^{2}-zx+x^{2}}+z \sqrt{x^{2}-x y+y^{2}} \leqslant x^2+y^2+z^2.$$
By the Cauchy-Schwarz inequality we have
$$\left(\sum x \sqrt{y^{2}-y z+z^{2}}\right) ^2 \leqslant (x+y+z) \sum x(y^2-yz+z^2).$$
It's remain to prove that
$$(x+y+z) \sum x(y^2-yz+z^2) \leqslant (x^2+y^2+z^2)^2,$$
equivalent to
$$x^4+y^4+z^4+xyz(x+y+z) \geqslant \sum xy(x^2+y^2).$$
Which is the Schur inequality.
