Interesting Inequality using AM-GM and other identities.

Question

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.

Could you please share in your post where you got stuck? Questions that show no attempt are usually closed. — Aaron Hendrickson, Jun 21 '21 at 23:05
@throwawayacc21 Thank you for the additional edits, +1 to your question. — Sarvesh Ravichandran Iyer, Jun 21 '21 at 23:42
Let $x=ab, y=bc, z=ca$, the last inequality is equivalent to $$\frac{5}{2}(xy+yz+zx) \le x^2+y^2+z^2+3xyz\left( \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \right)$$ which is wrong. Indeed, if you take $x=y=1,z \rightarrow 0$, you will have an wrong inequality, that is $ 2.5 \le 2$ — Paresseux Nguyen, Jun 22 '21 at 00:50
Even though the initial ineq is not really tight, it is not so loose that you can reason this way. — Paresseux Nguyen, Jun 22 '21 at 00:51

score 3 · Accepted Answer · answered Jun 22 '21 at 05:51

WLOG, assume $c = \max(a, b, c)$.

By AM-GM inequality, we have \begin{align*} \left(\frac{b^2}{c} - b + c\right) + c &\ge 2\sqrt{b^2 - bc + c^2}, \\ \left(\frac{a^2}{c} - a + c\right) + c &\ge 2\sqrt{a^2 - ac + c^2}, \\ \left(\frac{b^2}{a} - b + a\right) + a &\ge 2\sqrt{b^2 - ab + a^2}. \end{align*} It suffices to prove that $$\frac{b^2}{a} + \frac{a^2 + b^2}{c} + a - 2b + 4c \le 2\left(\frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b}\right)$$ that is $$\frac{(2c^2 - ab - bc)(a - b)^2}{abc} \ge 0$$ which is true.

We are done.

score 2 · Answer 2 · answered Jun 22 '21 at 05:50

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.

Interesting Inequality using AM-GM and other identities.

2 Answers2