4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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:

https://math.stackexchange.com/tags/uvw/info

and here:

https://artofproblemsolving.com/community/c6h278791

There is an easier proof:

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

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

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.