# Proving $\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$

Prove that (where $$a,b,c>0$$) $$\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$$

I have found a proof for this problem but it is very lengthy and is not nice.(Its not using computer though).I shall post it later.

Background: Here however is a very similar problem

If $$a,b,c>0$$ then prove that $$\sum_{cyc} \frac{a(a^2+bc)}{b+c}\ge a^2+b^2+c^2$$

There is a beautiful proof of this by Michael Rozenberg (arqady)

WLOG $$a\ge b\ge c$$ rewrite the inequality as $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\ge 0$$ but $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\geq\frac{a(a-b)(a-c)}{b+c}+\frac{b(b-a)(b-c)}{a+c}=$$ $$=(a-b)(\frac{a(a-c)}{b+c}-\frac{b(b-c)}{a+c})\geq0$$

However this method doesnt work...

I am looking for a clean and smooth proof (without using BW,uvw or complete expanding)

I am not planning to disclose how I proved the inequality as it may spoil the fun!

• What does cyc mean under the summation? Some kind of cyclic permutation? And if so of what? Commented Feb 11, 2021 at 17:53
• @RobertTheTutor The cyc mean the summation over cyclic pernutations $a\to b\to c$ for example $$\sum_{cyc} a+2b=a+2b+b+2c+c+2a$$ Commented Feb 11, 2021 at 17:55
• I am sure this can be done by SOS or SOS - Schur. Personally, I don't consider $uvw$ method less aesthetic than SOS. Commented Feb 11, 2021 at 23:22
• @AlbusDumbledore The desired inequality is written as $\sum_{\mathrm{cyc}} \frac{a^3}{b+c} + 2abc\sum_{\mathrm{cyc}}\frac{1}{b+c} \ge \frac{(a+b+c)^2}{2}$. The following is also true: $\sum_{\mathrm{cyc}} \frac{a^3}{b+c} + 2abc\frac{9}{2(a+b+c)} \ge \frac{(a+b+c)^2}{2}$ (C-S for the 2nd term in LHS). Commented Feb 12, 2021 at 2:13
• @AlbusDumbledore It is nice. Commented Feb 12, 2021 at 7:39

Suppose $$c = \min\{a,b,c\},$$ then $$\sum \frac{a^3+abc}{b+c}-(a^2+b^2+c^2) = \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-c)(b-a)}{c+a}+\frac{c(c-a)(c-b)}{a+b}$$ $$\geqslant \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-c)(b-a)}{c+a} = \frac{(a-b)^2(a^2+ab+b^2-c^2)}{(b+c)(c+a)} \geqslant 0.$$ Therefore $$\sum \frac{a^3+abc}{b+c} \geqslant a^2+b^2+c^2. \qquad (1)$$ Now, we write the inequality as $$\sum \frac{a^3+abc}{b+c} + abc \sum \frac{1}{b+c} \geqslant \frac{(a+b+c)^2}{2}.$$ By the Cauchy-Schwarz inequality, we have $$\sum \frac{1}{b+c} \geqslant \frac{9}{2(a+b+c)}. \qquad (2)$$ From $$(1)$$ and $$(2),$$ we need to prove $$a^2+b^2+c^2+\frac{9abc}{2(a+b+c)} \geqslant \frac{(a+b+c)^2}{2},$$ equivalent to $$a^2+b^2+c^2+\frac{9abc}{a+b+c} \geqslant 2(ab+bc+ca).$$ This is Schur inequality. The proof is completed.

• Very nice nguyenhuyen_ag!Nice to see that you used the inequality I had kept as background Bravo! Commented Feb 12, 2021 at 12:32
• @AlbusDumbledore Thank you =)) Commented Feb 12, 2021 at 13:05

We want to show $$\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$$

The following steps mainly show how this inequality can be maximally simplified. The final steps then have been proved by the OP Albus Dumbledore himself, and others.

Due to homogeneity, let $$a+b+c =1$$, then equivalently

$$\sum_{cyc} \frac{(a-1+1)(a^2+2bc)}{1-a} \ge \frac{1}{2}$$ $$-\sum_{cyc} (a^2+2bc) + \sum_{cyc} \frac{a^2+2bc}{1-a} \ge \frac{1}{2}$$ $$-(a+b+c)^2+ \sum_{cyc} \frac{a^2+2bc}{1-a} \ge \frac{1}{2}$$ $$\sum_{cyc} \frac{a^2-1+1+2bc}{1-a} \ge \frac{3}{2}$$ $$-\sum_{cyc} (1+a)+\sum_{cyc} \frac{1+2bc}{1-a} \ge \frac{3}{2}$$ $$\sum_{cyc} \frac{1+2bc}{b+c} \ge \frac{11}{2}$$ Now we have $$\frac{2bc}{1-a} =\frac{2bc}{b+c} = b+c -a-1 + \frac{1 -(a^2+b^2+c^2)}{b+c}$$ which leads to $$(2 -(a^2+b^2+c^2))\sum_{cyc} \frac{1}{b+c} \ge \frac{15}{2}$$ Now we have isolated a single sum, which is the main benefit of this answer.

This sum can be evaluated: $$\sum_{cyc} \frac{1}{1-a} = \frac{3 - (a^2+b^2+c^2)}{1 - 2abc - (a^2+b^2+c^2)}$$ which leaves to show $$(2 -(a^2+b^2+c^2))(3 -(a^2+b^2+c^2)) \ge \frac{15}{2} (1 - 2abc - (a^2+b^2+c^2))$$ Let $$x = a^2+b^2+c^2$$ then we have to show $$2x^2 + 5x + 30 abc \ge 3$$

Note that by now, no single change has been made to the original question, since all transformations are equivalences.

The last inequality can be proved (amongst other methods) by Schur, which has been done by the OP Albus Dumbledore himself, and others, see here. $$\qquad \Box$$

• +1 nice approach ,As I mentioned in question I had a proof ,I used the SOS method ,but this is slightly shorter and more elegant Thank you very much (I now understand the mostivation of that question too :) Commented Feb 12, 2021 at 17:04