Prove that for a + b + c =1 and a,b,c are positive real numbers, then

$$\frac{bc+a+1}{a^2+1} + \frac{ac+b+1}{b^2+1} + \frac{ab+c+1}{c^2+1} \le \frac{39}{10}$$

My try: if one term is proven to be $\le \frac{13}{10} $ then we prove the inequality. sub in $a =1-b-c$ then we have $$\frac{bc+b+c+2}{b^2+c^2+2bc-2b-2c+2}\le\frac{13}{10}$$ further simplifying, then we have $$0\le13b^2+13c^2+16bc-16b-16c+6$$ since $b^2+c^2\ge 2bc$ then we can substitute it? (not sure about this) then we have $$0\le42bc-16b-16c+6$$ since $b+c\ge2\sqrt{bc}$ then we can sub it again? (also not sure about this) then we get a quadratic inequality $$21bc -16\sqrt{bc} +6$$ but it is not true for all positive real numbers a, b and c.

Is it impossible to use this approach and get the answer (is there a better way to solve it) or did i do something wrong?

  • $\begingroup$ You are looking to prove that each term is $\geq \frac{13}{10}$? I am thinking that that goal is a bit too strong. For Olympiad style problems like this my first instinct is always AM GM, so maybe try using that to combine the terms? $\endgroup$ – Andrew Tindall May 6 '18 at 15:49
  • $\begingroup$ @AndrewTindall I was about to write the same and I tried $1,0,0$ in an attempt to prove that this approach cannot succeed. $\endgroup$ – Arnaud Mortier May 6 '18 at 15:51
  • 1
    $\begingroup$ Could it be that the inequality is the other way around? I tried some sample values and the LHS always turned out to be $\le 3.9$. $\endgroup$ – Martin R May 6 '18 at 15:51
  • $\begingroup$ Sorry i forgot that they are supposed tobe positive real numbers and should be the other way around $\endgroup$ – SuperMage1 May 6 '18 at 16:00
  • $\begingroup$ @SuperMage1: Then it still remains wrong for $b, c$ close to zero. $\endgroup$ – Martin R May 6 '18 at 16:04

We need to prove that $$\frac{39}{10}\geq\sum_{cyc}\frac{bc+a+1}{a^2+1}$$ or $$\frac{39}{10}\geq\sum_{cyc}\frac{bc+a(a+b+c)+1}{a^2+1}$$ or $$\frac{9}{10}\geq(ab+ac+bc)\sum_{cyc}\frac{1}{a^2+(a+b+c)^2}$$ or $$9\prod_{cyc}(a^2+(a+b+c)^2)\geq10(ab+ac+bc)\sum_{cyc}(a^2+(a+b+c)^2)(b^2+(a+b+c)^2),$$ for which it's enough to prove that

$$9\prod_{cyc}(a^2+(a+b+c)^2)\geq10(ab+ac+bc)\sum_{cyc}(a^2+(a+b+c)^2)(b^2+(a+b+c)^2)+$$ $$+\frac{1}{9}\left(\sum_{cyc}(a^3-a^2b-a^2c+abc)\right)^2,$$ which is true by uvw.

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

Now, since $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=$$ $$=27u^3-27uv^2+3w^3-9uv^2+3w^3+3w^3=9w^3+27u^3-36uv^2,$$ we see that the last inequality is a linear inequality of $w^3$,

which says that it's enough to prove the last inequality for an extreme value of $w^3$,

which happens for equality case of two variables.

Since the last inequality is homogeneous and even degree, it's enough to assume $b=c=1$, which gives $$9\left(a^2+(a+2)^2\right)\left(1+(a+2)^2\right)^2\geq10(2a+1)\left(2\left(a^2+(a+2)^2\right)(1+(a+2)^2)+\left(1+(a+2)^2\right)^2\right)+$$ $$+\frac{1}{9}(a^3+2-2(a^2+a+1)+3a)^2$$ or $$(a-1)^2(161a^4+1046a^3+2843a^2+3780a+2250)\geq0,$$ which is true because $$161a^4+1046a^3+2843a^2+3780a+2250=$$ $$=a^2(17a^2+14a+4)+(12a^2+43a+30)^2+30(9a^2+40a+45)>0.$$ Done!

A proof that the inequality $$9\prod_{cyc}(a^2+(a+b+c)^2)\geq10(ab+ac+bc)\sum_{cyc}(a^2+(a+b+c)^2)(b^2+(a+b+c)^2)+$$ $$+\frac{1}{9}\left(\sum_{cyc}(a^3-a^2b-a^2c+abc)\right)^2,$$ is a linear inequality of $w^3$. $$9\prod_{cyc}(a^2+(a+b+c)^2)=9\prod_{cyc}(a^2+9u^2)=9w^6+A(u,v^2)w^3+B(u,v^2),$$ $$10(ab+ac+bc)\sum_{cyc}(a^2+(a+b+c)^2)(b^2+(a+b+c)^2)=C(u,v^2)w^3+D(u,v^2)$$ and $$\frac{1}{9}\left(\sum_{cyc}(a^3-a^2b-a^2c+abc)\right)^2=$$ $$=\frac{1}{9}\left(27u^3-36uv^2+9w^3\right)^2=9w^6+E(u,v^2)w^3+F(u,v^2),$$ where $A$, $B$, $C$, $D$, $E$ and $F$ they are polynomials of $u$ and $v^2$ only.

We see that $9w^6$ canceled and we obtain a linear inequality of $w^3$.

About uvw see here:


and here:


There is an easier proof:

$$\frac{39}{10}-\sum_{cyc}\frac{bc+a+1}{a^2+1}=\sum_{cyc}\left(\frac{13}{10}-\frac{bc+a(a+b+c)+(a+b+c)^2}{a^2+(a+b+c)^2}\right)=$$ $$=\sum_{cyc}\frac{6a^2+3b^2+3c^2-4ab-4ac-4bc}{10(a^2+1)}=$$ $$=\sum_{cyc}\frac{(a-b)(3a-3b+2c)-(c-a)(3a-3c+2b)}{10(a^2+1)}=$$ $$=\sum_{cyc}(a-b)\left(\frac{3a-3b+2c}{10(a^2+1)}-\frac{3b-3a+2c}{10(b^2+1)}\right)=$$ $$=\sum_{cyc}\frac{(a-b)^2(3a^2+3b^2-2ac-2bc+6)}{10(a^2+1)(b^2+1)}\geq$$ $$\geq\sum_{cyc}\frac{(a-b)^2(\frac{3}{2}(a+b)^2-2(a+b)c+6)}{10(a^2+1)(b^2+1)}=$$ $$=\sum_{cyc}\frac{(a-b)^2(3(1-c)^2-4(1-c)c+12)}{20(a^2+1)(b^2+1)}=\sum_{cyc}\frac{(a-b)^2(7c^2-10c+15)}{20(a^2+1)(b^2+1)}\geq0.$$

  • $\begingroup$ Can you explain how you recognized the "now since" part . And why being a linear inequality of $w^3$ means it is enough to prove for an extreme value of $w^3$? (I think $uv^2 > w^3$ so wouldn't you want $w^3$ to be small??) $\endgroup$ – user106860 May 7 '18 at 0:48
  • $\begingroup$ @user106860 I added something. See now. $\endgroup$ – Michael Rozenberg May 7 '18 at 4:21
  • $\begingroup$ sorry, but could you explain the inequality after"for which it is enough to prove that". $\endgroup$ – SuperMage1 May 7 '18 at 12:06
  • $\begingroup$ I explained that this inequality is a linear inequality of $w^3$. Is this clear? $\endgroup$ – Michael Rozenberg May 7 '18 at 13:07
  • $\begingroup$ @SuperMage1There is an easier proof. I added this proof. See now. $\endgroup$ – Michael Rozenberg May 7 '18 at 16:10

This is wrong. If $a=1$ and $b=c=0$ then the left-hand side is $3$, which is less than $\frac{39}{10}$.

  • $\begingroup$ Already edited ^^ $\endgroup$ – SuperMage1 May 6 '18 at 16:07

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