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Let $x,y,z>0$, be such that $x+y+z=3$. Show that $$\dfrac{x^2}{2y^2-y+3}+\dfrac{y^2}{2z^2-z+3}+\dfrac{z^2}{2x^2-x+3}\ge\dfrac{3}{4}.$$

I've tried many things but all have failed. $$\left(\sum_{\text{cyc}}\dfrac{x^2}{2y^2-y+3}\right)\left(\sum_{\text{cyc}}(2y^2-y+3)\right)\ge (x+y+z)^2=9.$$ But $$\sum_{\text{cyc}}(2y^2-y+3)=2\sum_{\text{cyc}}x^2+6\ge 12.$$

3 Answers3

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Write $x= \frac{3a}{a+b+c}$, $y=\frac{3b}{a+b+c}$, $z=\frac{3c}{a+b+c}$ and substitute. Get the equivalent inequality:

\begin{eqnarray} 18 (a^6 +b^6+c^6) + 33 (a^5 b +b^5 c+ c^5a ) + 2 (a^4 b^2+ b^4 c^2 + c^4 a^2)-50(a^3 b^3 + b^3 c^3 + c^3 a^3)+\\ + 98 (a^2 b^4 + b^2 c^4 + c^2 a^4) + 9 (a b^5 + b c^5 + c a^5) - 6 (a^4 b c + b^4 c a + c^4 a b )-\\ - 11( a^3 b^2 c + b^3 c^2 a + c^3 a^2 b)- 35(a^3 b c^2 + b^3 c a^2 + c^3 a b^2)- 174 a^2 b^2c^2 \ge 0 \end{eqnarray} with $a$, $b$, $c >0$. The expression on the left hand side has circular symmetry. So we may assume that either $a\ge b\ge c >0$ or $0< a \le b \le c$. In the first case, substitute $c= u$, $b = u+v$, $a= u+v+w$ in the above expression and we get an expression in $u$, $v$, $w$ with all the coefficients positive, hence $\ge 0$. Similarly, in the second case use $a= u$, $b=u+v$, $c=u+v+w$, again getting on the LHS an expression in $u$,$v$,$w$ with all the coefficients positive.

$\bf{Added:}$

When performing the substitution $c= u$, $b = u+v$, $a= u+v+w$ we get $$696 u^4 v^2 + 1774 u^3 v^3 + 1720 u^2 v^4 + 752 u v^5 + 128 v^6 + 696 u^4 v w + 2409 u^3 v^2 w + 2936 u^2 v^3 w + 1592 u v^4 w + 336 v^5 w + 696 u^4 w^2 + 2655 u^3 v w^2 + 3627 u^2 v^2 w^2 + 2226 u v^3 w^2 + 560 v^4 w^2 + 1010 u^3 w^3 + 2411 u^2 v w^3 + 1999 u v^2 w^3 + 648 v^3 w^3 + 574 u^2 w^4 + 913 u v w^4 + 437 v^2 w^4 + 150 u w^5 + 141 v w^5 + 18 w^6$$

while performing $a= u$, $b=u+v$, $c=u+v+w$ we get $$696 u^4 v^2 + 1774 u^3 v^3 + 1720 u^2 v^4 + 752 u v^5 + 128 v^6 + 696 u^4 v w + 2913 u^3 v^2 w + 3944 u^2 v^3 w + 2168 u v^4 w + 432 v^5 w + 696 u^4 w^2 + 3159 u^3 v w^2 + 5139 u^2 v^2 w^2 + 3378 u v^3 w^2 + 800 v^4 w^2 + 1010 u^3 w^3 + 2915 u^2 v w^3 + 2647 u v^2 w^3 + 792 v^3 w^3 + 574 u^2 w^4 + 985 u v w^4 + 413 v^2 w^4 + 150 u w^5 + 117 v w^5 + 18 w^6$$

We see that we have equality if and only if $v=w=0$, that is, $a=b=c$, or, $x=y=z=1$.

orangeskid
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  • What are the expressions? – marty cohen Jul 12 '15 at 06:44
  • @marty cohen: Added some details, glad you asked. – orangeskid Jul 12 '15 at 07:37
  • Nice answer! Could you please explain how you came up with your original substitutions for x, y, z? – user84413 Jul 12 '15 at 18:39
  • That's a standard technique. If 0 < a < b < c, then you can write b = a+u, c = b+v = a+u+v where u > 0, v > 0. A multinomial in a, b, c then becomes a multinomial in a, u, v. If this multinomial has all non-negative coefficients, then it is always positive. – marty cohen Jul 12 '15 at 19:47
  • @user84413: Yes, the sum of $x$,$y$, $ z$ is $3$ so they are triple of some numbers with sum $1$. Now, for the numbers with sum $1$, they are some proportions. – orangeskid Jul 13 '15 at 20:53
  • @orangeskid Thanks for your reply; you taught me something new. – user84413 Jul 13 '15 at 21:35
  • @martycohen Thanks for your comment. (That part of the solution made sense, but I didn't know how to get the substitutions at the very start.) – user84413 Jul 13 '15 at 21:37
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By C-S $\sum\limits_{cyc}\frac{x^2}{2y^2-y+3}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(2x^2y^2-x^2y+3x^2)}$. Thus, it remains to prove that $\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(2x^2y^2-x^2y+3x^2)}\geq\frac{3}{4}$, which is $\sum\limits_{cyc}(3x^4-x^3y-2x^3z+x^2y^2-x^2yz)\geq0$, which is obvious.

Michael Rozenberg
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I would go into homogeneous coordinates, $$A=3=x+y+z$$ $$X=2x-y-z \quad x=(A+X)/3$$ and cyclically for $Y$ and $Z$. The coordinates $X$, $Y$, $Z$ are orthogonal to $A$, so whatever their values, they respect the constraint to the plane. Additionally, $X+Y+Z=0$.

With this, the sum is rewritten: $$\sum \frac{(A+X)^2}{2(A+Y)^2-3(A+Y)+27}=\sum\frac{9+6X+X^2}{36+3Y+2Y^2}$$ The $X=Y=Z=0$ lies on the symmetry axis, and there, the sum is exactly $\frac{3}{4}$. We now just have to prove that the function is increasing off-axis.

The denominator has no poles on the domain, so the Taylor expansion converges. Just write

$$\frac{1}{4}\sum (1+X/2+X^2/9)(1-(Y/12+Y^2/18)+(Y/12+Y^2/18)^2-(Y/12+Y^2/18)^3+\cdots)$$

and you can extract the second derivatives quite easily.

It may be a bit tricker to prove the function does not turn around and reach below the axis value, but at least you know where to start.

orion
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  • I like the use of homogenous coordinates. Somehow you need to use that $x$, $y$,.. are $>0$. – orangeskid Jul 11 '15 at 08:58
  • Perhaps you can write $x = \frac{3a}{a+b+c}$ and so on, and get a homogenous inequality in positive $a$, $b$, $c$ (also with circular symmetry). – orangeskid Jul 11 '15 at 09:06