There is the following problem.

Let $a$, $b$ and $c$ be real numbers such that $\prod\limits_{cyc}(a+b)\neq0$ and $k\geq2$ such that $\sum\limits_{cyc}(a^2+kab)\geq0.$ Prove that: $$\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}\geq\frac{9}{4}.$$

I have a proof of this inequality for any $k\geq2.6$.

I think, for $k<2.6$ it's wrong, but my software does not give me a counterexample

and I don't know, how to prove it for some $k<2.6$.

It's interesting that without condition $\sum\limits_{cyc}(a^2+kab)\geq0$ the equality occurs also for $(a,b,c)=(1,1,-1)$.

My question is: What is a minimal value of $k$, for which this inequality is true?

Thank you!


We want to show that your inequality does not hold for $k\in[2,13/5)$. In view of the identity in your answer, it is enough to show that for each $k\in[2,13/5)$ there is a triple $(a,b,c)\in\mathbb R^3$ with the following properties: $a=-1>b$, \begin{align}s_4&:=\sum_{cyc}(2a^3-a^2b-a^2c) \\ &=a^2 (2 a-b-c)+b^2 (-a+2 b-c)+c^2 (-a-b+2 c)=0, \end{align} $$s_3:=a b + b c + c a<0,$$ and $$s_2+k s_3=0,$$ where $$s_2:=a^2 + b^2 + c^2.$$ Indeed, then the right-hand side of your identity will be $$\frac{20}{3}\sum_{cyc}(a^4-a^2b^2)(13/5-k)s_3<0,$$ so that your identity will yield $$\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}<9/4.$$

For each $k\in(2,13/5)$, the triple $(a,b,c)$ will have all the mentioned properties if $a=-1$, $b$ is the smallest (say) of the 6 real roots $x$ of the polynomial $$P_k(x):=-18 - 15 k + 4 k^2 + 4 k^3 + (36 k + 6 k^2 - 12 k^3) x + (-27 - 9 k - 21 k^2 + 6 k^3) x^2 + (18 + 60 k - 10 k^2 + 8 k^3) x^3 + (-27 - 9 k - 21 k^2 + 6 k^3) x^4 + (36 k + 6 k^2 - 12 k^3) x^5 + (-18 - 15 k + 4 k^2 + 4 k^3) x^6, $$ and $$c=\tfrac12\, (k - b k) - \tfrac12\, \sqrt{-4 - 4 b^2 + 4 b k + k^2 - 2 b k^2 + b^2 k^2}.$$ For $k=2$, $(a,b,c)=(-1,0,1)$ will be such a triple.

So, we are done.

This result was obtained with Mathematica, as follows (which took Mathematica about 0.05 sec):

enter image description here

  • $\begingroup$ Thank you very much! $\endgroup$ – Michael Rozenberg Apr 21 '20 at 19:11

Your inequality fails to hold when e.g. $k=25999/10000=2.6-10^{-4}$ and $(a,b,c)=(97661/65536,-5/3,-1)$.

Indeed, the smallest value for which your inequality holds is $13/5=2.6$. Here is a proof by Mathematica:

enter image description here

enter image description here

So, the value $13/5$ of $k$ is witnessed by $$a=-1,\ b=x_*,\ c=\frac{1}{10} \left(13-13 x_*-\sqrt{69 x_*^2-78 x_*+69}\right),\tag{1}$$ where $x_*=-1.68\ldots$ is the smallest root of the $6$ real roots of the polynomial $$p(x)=1681 - 3198 x - 3621 x^2 + 10292 x^3 - 3621 x^4 - 3198 x^5 + 1681 x^6.$$

For this proof, Mathematica took about 32 sec, which is a huge time for a computer.

The values of the sums $$(s_1,s_2,s_3):=\left(\frac{2 a^2+b c}{(b+c)^2}+\frac{2b^2+a c}{(a+c)^2}+\frac{2 c^2+a b}{(a+b)^2},a^2+b^2+c^2,a b+a c+b c\right)$$ for the extremal $(a,b,c)$ given by (1) are
$$\Big(\frac94,-\frac{13}5\,s_{3*},s_{3*}\Big),$$ where $s_{3*}=-2.34\ldots$ is the smallest root of the $3$ real roots of the polynomial $$p_3(x)=3375 + 8775 x + 7065 x^2 + 1681 x^3,$$ with the other two roots $-1.04\ldots$ and $-0.826\ldots$.

  • $\begingroup$ Thank you, @Iosif ! It says that I was right. Now, what is a minimal value of $k$ for which this inequality is true? $\endgroup$ – Michael Rozenberg Apr 20 '20 at 18:27
  • $\begingroup$ I have a proof for any $k\geq\frac{13}{5}.$ I think it's a bug of the Mathematica. $\endgroup$ – Michael Rozenberg Apr 20 '20 at 18:49
  • $\begingroup$ @MichaelRozenberg : You are probably right. $\endgroup$ – Iosif Pinelis Apr 20 '20 at 18:53
  • $\begingroup$ Wow, @Iosif Pinelis ! I have no possibility to check these things.I make most of the computations by hand. It seems that $\frac{13}{5}$ is indeed a minimal value. Now, how we can prove it? $\endgroup$ – Michael Rozenberg Apr 20 '20 at 18:58
  • $\begingroup$ I have found a solution of my inequality for $k=\frac{13}{5}$ in one line!!! My proof was very complicated before. $\endgroup$ – Michael Rozenberg Apr 21 '20 at 0:26

I have found the following identity, which solves my problem for $k=\frac{13}{5}.$ $$4\prod_{cyc}(a+b)^2\left(\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}-\frac{9}{4}\right)=\frac{1}{3}\left(\sum_{cyc}(2a^3-a^2b-a^2c)\right)^2+$$ $$+\frac{20}{3}\sum_{cyc}(a^4-a^2b^2)\sum_{cyc}\left(a^2+\frac{13}{5}ab\right).$$ I got this identity by the following reasoning.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $\mathbb{w}(p)$ be a coefficient before $w^6$ in writing of a symmetric polynomial $p$ of three variables $a$, $b$ and $c$ as a polynomial of $u$, $v^2$ and $w^3$.

Thus, $$\mathbb{w}\left(4\prod_{cyc}(a+b)^2\left(\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}-\frac{9}{4}\right)\right)=$$ $$=\mathbb{w}\left(4\sum_{cyc}(2a^2+bc)(a^2+3v^2)^2-9(9uv^2-w^3)^2\right)=$$ $$=\mathbb{w}(8(a^6+b^6+c^6)+4abc(a^3+b^3+c^3)-9w^6)=24+12-9=27.$$ Now, we'll choose $m$, $n$ and $k$ such that the inequality $$4\prod_{cyc}(a+b)^2\left(\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}-\frac{9}{4}\right)\geq$$ $$\geq\frac{1}{3m^2}\left(\sum\limits_{cyc}(a^3+m(a^2b+a^2c)-(2m+1)abc)\right)^2+$$ $$+n\sum_{cyc}(a^4-a^2b^2)\sum_{cyc}(a^2+kab)$$ would be true for any reals $a$, $b$ and $c$ such that $\sum\limits_{cyc}(a^2+kab)\geq0.$

Indeed, since $$\mathbb{w}\left(\frac{1}{3m^2}\left(\sum\limits_{cyc}(a^3+m(a^2b+a^2c)-(2m+1)abc)\right)^2\right)=$$ $$=\mathbb{w}\left(\frac{1}{3m^2}(3w^3-3mw^3-3(2m+1)w^3)^2\right)=27,$$ we see that the last inequality is a linear inequality of $w^3$,

which by $uvw$ (see here: https://artofproblemsolving.com/community/c6h278791 ) says that it's enough to assume $b=c=1$ (the case $b=c=0$ we can check later).

Also, since for $b=c=1$ we have $$4\prod_{cyc}(a+b)^2\left(\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}-\frac{9}{4}\right)=8(a-1)^2(a+1)^2(a+2)^2$$ and $$\sum_{cyc}(a^4-a^2b^2)=(a^2-1)^2,$$ we see that $$\sum\limits_{cyc}(a^3+m(a^2b+a^2c)-(2m+1)abc)$$ has for $b=c=1$ a factor $a+1$, which gives $m=-\frac{1}{2}$ and we obtain: $$8(a+1)^2(a-1)^2(a+2)^2\geq$$ $$\geq\frac{4}{3}(a-1)^4(a+1)^2+n(a-1)^2(a+1)^2(a^2+2+k(2a+1))$$ or $$(20-3n)a^2+(104-6kn)a+92-3n(k+2)\geq0,$$ for which we need $$20-3n\geq0$$ and $$(52-3kn)^2-(20-3n)(92-3n(k+2))=0,$$ where the last it's $$(4+2n-kn)(kn+n-24)=0.$$ Here both cases give a same result.

For example, $$n=\frac{24}{k+1}$$ gives $$20-\frac{72}{k+1}\geq0$$ or $$k\geq\frac{13}{5}.$$ For $k=\frac{13}{5}$ we obtain $n=\frac{20}{3}$ and for these values of $n$ and $k$ our inequality turned out an identity!

  • 6
    $\begingroup$ Yes, I'd certainly like to see how this identity was found. $\endgroup$ – Iosif Pinelis Apr 21 '20 at 13:01
  • 1
    $\begingroup$ This is amazing arqady sir. Your work gives inspiration. $\endgroup$ – Aditya Guha Roy Jun 23 '20 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.