Prove $\Big[\sum\limits_{cyc} a(a^2+2bc)\Big]^3 \geqq 3(ab+bc+ca)^2 . \sum\limits_{cyc} a(a^2+2bc)^2$

Question

For $a,b,c>0$, prove that: $$ \Big[\sum\limits_{cyc} a(a^2+2bc)\Big]^3 \geqq 3(ab+bc+ca)^2 . \sum\limits_{cyc} a(a^2+2bc)^2$$ BW works here, but it's very ugly!

My try: Let $p=a+b+c,q=ab+bc+ca,r=abc$. We need to prove: $${p}^{9}-9\,{p}^{7}q+27\,{p}^{6}r+24\,{p}^{5}{q}^{2}-162\,{p}^{4}qr-12 \,{p}^{3}{q}^{3}+243\,{p}^{3}{r}^{2}+216\,{p}^{2}{q}^{2}r-15\,p{q}^{4} -729\,pq{r}^{2}+27\,{q}^{3}r+729\,{r}^{3} \geqq 0$$ However, I don't know what I need to do next? Another work:

Assume $c=\min\{a,b,c\}$ and $f(a,b,c) =\text{LHS-RHS}$. First, we prove: $$f(a,b,c) \geqq f(\frac{a+b}{2},\frac{a+b}{2},c) \Leftarrow \frac{3}{256} (a-b)^2 M \geqq 0$$ Thus, we need to prove: $M\geqq 0$, which is easy for $c=\min\{a,b,c\}$ but very ugly!

Now we prove: $$f(\frac{a+b}{2},\frac{a+b}{2},c) \geqq 0$$

Or $${ \left( {a}^{7}+7\,{a}^{6}b+16\,{a}^{6}c+21\,{a}^{5}{b}^{2}+96 \,{a}^{5}bc+108\,{a}^{5}{c}^{2}+35\,{a}^{4}{b}^{3}+240\,{a}^{4}{b}^{2} c+540\,{a}^{4}b{c}^{2}+272\,{a}^{4}{c}^{3}+35\,{a}^{3}{b}^{4}+320\,{a} ^{3}{b}^{3}c+1080\,{a}^{3}{b}^{2}{c}^{2}+1088\,{a}^{3}b{c}^{3}+80\,{a} ^{3}{c}^{4}+21\,{a}^{2}{b}^{5}+240\,{a}^{2}{b}^{4}c+1080\,{a}^{2}{b}^{ 3}{c}^{2}+1632\,{a}^{2}{b}^{2}{c}^{3}+240\,{a}^{2}b{c}^{4}+144\,{a}^{2 }{c}^{5}+7\,a{b}^{6}+96\,a{b}^{5}c+540\,a{b}^{4}{c}^{2}+1088\,a{b}^{3} {c}^{3}+240\,a{b}^{2}{c}^{4}+288\,ab{c}^{5}+64\,a{c}^{6}+{b}^{7}+16\,{ b}^{6}c+108\,{b}^{5}{c}^{2}+272\,{b}^{4}{c}^{3}+80\,{b}^{3}{c}^{4}+144 \,{b}^{2}{c}^{5}+64\,b{c}^{6}+64\,{c}^{7} \right) \left( a+b-2\,c \right) ^{2}}\geqq 0$$

So on, I think it's hard to find a nice proof for it? Without ”Tejs’s Theorem” in uvw?

PS: The original inequality is https://artofproblemsolving.com/community/c6h2080774p15009613

Answer 1

I think, just $uvw$ gives a nice solution.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $$(a^3+b^3+c^3+6abc)^3\geq3(ab+ac+bc)^2\sum_{cyc}(a^5+4a^3bc+4a^2b^2c)$$ or $$(27u^3-27uv^2+3w^3+6w^3)^3\geq$$ $$\geq27v^4(243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3+36u^2w^3-24v^2w^3+12v^2w^3)$$ or $f(w^3)\geq0,$ where $$f(w^3)=(3u^3-3uv^2+w^3)^3-v^4(9u^5-15u^3v^2+5uv^4+3u^2w^3-v^2w^3).$$ But by Schur $$f'(w^3)=3(3u^3-3uv^2+w^3)^2-v^4(3u^2-v^2)\geq$$ $$\geq3(3u^3-3uv^2+4uv^2-3u^3)^2-v^4(3u^2-v^2)=v^6>0,$$ which says that $f$ increases.

Thus, by $uvw$ (https://artofproblemsolving.com/community/c6h278791 )

it's enough to prove our inequality in two cases:

1) $w^3\rightarrow0^+$;

2) Two variables are equal.

Can you end it now?

In both cases we obtain right inequalities.

It seems that the original inequality, from which comes your problem, has really nice proof.

Answer 2

The original problem we can prove by Minkowski and SOS.

Indeed, we need to prove that: $$\sum_{cyc}a\sqrt{a^2+2bc}\geq\sqrt3(ab+ac+bc)$$ for non-negatives $a$, $b$ and $c$ or $$\sum_{cyc}\sqrt{a^8+2a^4b^2c^2}\geq\sqrt3\sum_{cyc}a^2b^2.$$ Now, by Minkowski $$\sum_{cyc}\sqrt{a^8+2a^4b^2c^2}\geq\sqrt{\left(\sum_{cyc}a^4\right)^2+2a^2b^2c^2\left(\sum_{cyc}a\right)^2}.$$ Thus, it's enough to prove that: $$\left(\sum_{cyc}a^4\right)^2+2a^2b^2c^2\left(\sum_{cyc}a\right)^2\geq3\left(\sum_{cyc}a^2b^2\right)^2$$ or $$\sum_{cyc}(a^8-a^4b^4-4a^4b^2c^2+4a^3b^3c^2)\geq0$$ or $$\sum_{cyc}\left(\frac{1}{2}(a^4-b^4)^2-2a^2b^2c^2(a-b)^2\right)\geq0$$ or $$\sum_{cyc}(a-b)^2((a+b)^2(a^2+b^2)^2-4a^2b^2c^2)\geq0$$ or $$\sum_{cyc}(a-b)^2((a+b)(a^2+b^2)+2abc)((a+b)(a^2+b^2)-2abc)\geq0$$ and since $$a^2+b^2\geq2ab,$$ it's enough to prove that $$\sum_{cyc}(a-b)^2((a+b)(a^2+b^2)+2abc)ab(a+b)(a+b-c)\geq0.$$ Now, let $a\geq b\geq c$.

Thus, $$\sum_{cyc}(a-b)^2((a+b)(a^2+b^2)+2abc)ab(a+b)(a+b-c)\geq$$ $$\geq (a-c)^2((a+c)(a^2+c^2)+2abc)ac(a+c)(a+c-b)+$$ $$+(b-c)^2((b+c)(b^2+c^2)+2abc)bc(b+c)(b+c-a)\geq$$ $$\geq (b-c)^2((a+c)(a^2+c^2)+2abc)ac(a+c)(a-b)+$$ $$+(b-c)^2((b+c)(b^2+c^2)+2abc)bc(b+c)(b-a)=$$ $$=(b-c)^2(a-b)c(a((a+c)(a^2+c^2)+2abc)(a+c)-b((b+c)(b^2+c^2)+2abc)(b+c))\geq0.$$

Answer 3

I think a nicest proof gives the following Holder. $$\left(\sum_{cyc}a\sqrt{a^2+2bc}\right)^2\sum_{cyc}\frac{a}{a^2+2bc}\geq(a+b+c)^3.$$ Thus, it's enough to prove that $$\frac{(a+b+c)^3}{3(ab+ac+bc)^2}\geq\sum_{cyc}\frac{a}{a^2+2bc}$$ and since $$(a+b+c)^2\geq3(ab+ac+bc),$$ it's enough to prove that $$\frac{a+b+c}{ab+ac+bc}\geq\sum_{cyc}\frac{a}{a^2+2bc}$$ or $$\sum_{cyc}\left(\frac{a}{ab+ac+bc}-\frac{a}{a^2+2bc}\right)\geq0$$ or $$\sum_{cyc}\frac{a(a-b)(a-c)}{a^2+2bc}\geq0.$$ Now, let $a\geq b\geq c$.

Thus, $$b\sum_{cyc}\frac{a(a-b)(a-c)}{a^2+2bc}\geq\frac{a(a-b)b(a-c)}{a^2+2bc}+\frac{b^2(b-a)(b-c)}{b^2+2ac}\geq$$ $$\geq\frac{a(a-b)a(b-c)}{a^2+2bc}+\frac{b^2(b-a)(b-c)}{b^2+2ac}=$$ $$=(a-b)(b-c)\left(\frac{a^2}{a^2+2bc}-\frac{b^2}{b^2+2ac}\right)\geq0.$$