# Math Help - Inequality

1. ## Inequality

Prove that if $ab+bc+ca=\frac13$ ( $a,b,c$ are positive reals) then
$\frac a{a^2-bc+1}+\frac b{b^2-ca+1}+\frac c{c^2-ab+1}\ge \frac 1{a+b+c}$.

2. I am sure if you multiplied the inequality by the LCD then that may help.

3. This problem looks similar to this one. Is the method to be used similar?

4. Probably. (I haven't done it so don't know but it's from a series...) Maybe the inequality could be used as we proved it.

5. This problem has been gripping me for a while now and still I can’t solve it. Here’s what I’ve done with the inequality so far. Multiply through by a+b+c:

$\sum{\frac{a^2+ab+ca}{a^2-bc+1}}\ \geq\ 1$

$\sum{\frac{a^2-bc+\frac{1}{3}}{a^2-bc+1}}\ \geq\ 1$

$\sum{\left[1-\frac{2}{3}\,\left(\frac{1}{a^2-bc+1}\right)\right]}\ \geq\ 1$

$\therefore\ \frac{1}{a^2-bc+1}+\frac{1}{b^2-ca+1}+\frac{1}{c^2-ab+1}\ \leq\ 3$

Now

$\frac{1}{a^2-bc+1}\ =\ \frac{1}{a^2+ab+ca+\frac{2}{3}}\ =\ \frac{3}{3a(a+b+c)+2}$

Thus

$\frac{1}{3a(a+b+c)+2}+\frac{1}{3b(a+b+c)+2}+\frac{ 1}{3c(a+b+c)+2}\ \leq\ 1$

Let $A=\sqrt{3}a,\ B=\sqrt{3}b,\ C=\sqrt{3}c$. Then $AB+BC+CA=1$ and the inequality becomes

$\color{white}.\hspace{15mm}\color{black}\frac{1}{A (A+B+C)+2}+\frac{1}{B(A+B+C)+2}+\frac{1}{C(A+B+C)+ 2}\ \leq\ 1$

This is what needs to be proved. I think the trick is to let $A=\tan{\alpha},\ B=\tan{\beta},\ C=\tan{\gamma}$, where $\alpha+\beta+\gamma=\frac{\pi}{2}$, but I still can’t do it.

6. This problem has also got my head interacting with brick walls.

Following JaneBennet's ideas, we want to show that if A, B, C are positive numbers with BC+CA+AB=1, then

$\color{white}.\hspace{15mm}\color{black}\frac{1}{A (A+B+C)+2} + \frac{1}{B(A+B+C)+2} + \frac{1}{C(A+B+C)+2}\ \leq\ 1$.

Let $x^3 - px^2 + qx -r = 0$
be the cubic equation with roots A, B and C. Then p=A+B+C, q=BC+CA+AB and r=ABC. Hence q=1.

The equation $(x/p)^3 - p(x/p)^2 + (x/p) - r = 0$ has roots Ap, Bp and Cp. In other words, the equation $x^3 - p^2x^2 + p^2x - p^3r = 0$ has roots A(A+B+C), B(A+B+C), C(A+B+C).

Therefore the equation $(x-2)^3 - p^2(x-2)^2 + p^2(x-2) - p^3r = 0$ has roots A(A+B+C)+2, B(A+B+C)+2, C(A+B+C)+2. In other words (multiplying out the brackets), the equation $x^3 - (p^2+6)x^2 + (5p^2+12)x - (p^3r+6p^2+8) = 0$ has those roots.

Replacing x by 1/x, we see that the equation $(p^3r+6p^2+8)x^3 - (5p^2+12)x^2 + (p^2+6)x - 1 = 0$ has roots $\frac1{A(A+B+C)+2}$, $\frac1{B(A+B+C)+2}$, $\frac1{C(A+B+C)+2}$.

The sum of the roots of this cubic equation is also the coefficient of x^2 divided by the coefficient of x^3. So to prove our inequality we need to show that $\frac{5p^2+12}{p^3r+6p^2+8}\leqslant1$, where p=A+B+C and r=ABC. An equivalent inequality is $(rp+1)p^2\geqslant4$.

Now what?

7. Originally Posted by JaneBennet
This problem has been gripping me for a while now and still I can’t solve it. Here’s what I’ve done with the inequality so far. Multiply through by a+b+c:

$\sum{\frac{a^2+ab+ca}{a^2-bc+1}}\ \geq\ 1$

$\sum{\frac{a^2-bc+\frac{1}{3}}{a^2-bc+1}}\ \geq\ 1$

$\sum{\left[1-\frac{2}{3}\,\left(\frac{1}{a^2-bc+1}\right)\right]}\ \geq\ 1$

$\therefore\ \frac{1}{a^2-bc+1}+\frac{1}{b^2-ca+1}+\frac{1}{c^2-ab+1}\ \leq\ 3$

Now

$\frac{1}{a^2-bc+1}\ =\ \frac{1}{a^2+ab+ca+\frac{2}{3}}\ =\ \frac{3}{3a(a+b+c)+2}$

Thus

$\frac{1}{3a(a+b+c)+2}+\frac{1}{3b(a+b+c)+2}+\frac{ 1}{3c(a+b+c)+2}\ \leq\ 1$

I can show a+b+c > 1 ,and hence we will be required to show,
$\frac{1}{3a+2}+\frac{1}{3b+2}+\frac{1}{3c+2}\ \leq\ 1$

8. Originally Posted by Isomorphism
I can show a+b+c > 1 ,and hence we will be required to show,
$\frac{1}{3a+2}+\frac{1}{3b+2}+\frac{1}{3c+2}\ \leq\ 1$
Unfortunately that inequality need not hold. For example, if a=0, b=c=1/√3, then bc+ca+ab=1, but $\frac{1}{3a+2}+\frac{1}{3b+2}+\frac{1}{3c+2}>1.03$.

9. Originally Posted by Opalg
Unfortunately that inequality need not hold. For example, if a=0, b=c=1/√3, then bc+ca+ab=1, but $\frac{1}{3a+2}+\frac{1}{3b+2}+\frac{1}{3c+2}>1.03$.
ya thats true, that means it's a strict inequality. Hmm i am trying it from a long time, but no use

10. Okay, I can complete the proof of this, but it's not at all illuminating. There must be a more conceptual explanation for this strange inequality.

The situation is that A, B, C are positive numbers with BC+CA+AB=1. If p=A+B+C and r=ABC then we want to show that $rp^3+p^2\geqslant4$. Since the proof involves calculus, I'll change the notation (again!) and write x, y, z for A, B, C.

So x, y, z are positive reals with yz+zx+xy=1, and we want to show that $f(x,y,z) := xyz(x+y+z)^3 + (x+y+z)^2\geqslant4$. Let $D=\{(x,y,z)\in\mathbb{R}^3:x,y,z\geqslant0, yz+zx+xy=1\}$. Then we want to show that the minimum value of f on D is 4.

The boundary of D consists of points for which one of the coordinates is zero. Then the other two coordinates must be w and 1/w, for some positive number w. The value of f at this point is $(w+w^{-1})^2$, and it is easy to see that the minimum of this expression is 4 (occurring when w=1).

To find the extrema of f in the interior of D, use the method of Lagrange multipliers, and find the zeros of the partial derivatives of g(x,y,z)=f(x,y,z)+λ(yz+zx+xy-1):
$\frac{\partial g}{\partial x} = yz(x+y+z)^3 + 3xyz(x+y+z)^2 +2(x+y+z) + \lambda(y+z) = 0$ ,
$\frac{\partial g}{\partial y} = zx(x+y+z)^3 + 3xyz(x+y+z)^2 +2(x+y+z) + \lambda(z+x) = 0$ ,
$\frac{\partial g}{\partial z} = xy(x+y+z)^3 + 3xyz(x+y+z)^2 +2(x+y+z) + \lambda(x+y) = 0$ .

Subtract the last of these from the middle one, to get $(z-y)\{x(x+y+z)^3 + \lambda\} = 0$ (and two similar equations with x,y and z cyclically interchanged). If x,y and z are all different then $\lambda = -x(x+y+z)^3 = -y(x+y+z)^3 = -z(x+y+z)^3$, which is impossible (since it implies that x, y and z are all equal).

Therefore the extreme values of f on D must occur at points where at least two of x, y and z are equal. Suppose for example that z=y. Then the conditon (x,y,z)∈D becomes y(2x+y)=1, and so $x={\textstyle\frac12}(y^{-1}-y)$. Note that this implies $0 (otherwise x would be negative). Also, $x+y+y = \frac{1+3y^2}{2y}$ and $xy^2 = {\textstyle\frac12}y(1-y^2)$.

Then $f(x,y,y) = {\textstyle\frac12}y(1-y^2)\frac{(1+3y^2)^3}{(2y)^3} + \frac{(1+3y^2)^2}{(2y)^2}$, which simplifies(?) to $\frac{5+32y^2+54y^4-27y^8}{16y^2}$. We want this to be greater than or equal to 4, or in other words $5+32y^2+54y^4-27y^8\geqslant64y^2$. When the term 64y^2 is taken over to the left side, this factorises as $(1-y^2)(1-3y^2)^2(5+3y^2)\geqslant0$, and this is clearly true when 0<y≤1.

11. Actually, it can be shown that $p^2\geq3$. By the Cauchy–Schwarz inequality,

$\sqrt{A^2+B^2+C^2}\sqrt{B^2+C^2+A^2}\ \geq\ AB+BC+CA\$

$\Rightarrow\ A^2+B^2+C^2 \geq\ 1$

$\Rightarrow\ (A+B+C)^2-2(AB+BC+CA)\ \geq\ 1$

$\Rightarrow\ (A+B+C)^2\ \geq\ 1+2\ =\ 3$

So it would appear to be simpler to prove that $rp^3\geq1$. The trouble is that $r\ngeq\frac{1}{3\sqrt{3}}$; in fact, the inequality is the other way round: $ABC\leq\frac{1}{3\sqrt{3}}$.

12. Originally Posted by JaneBennet
So it would appear to be simpler to prove that $rp^3\geq1$. The trouble is that $r\ngeq\frac{1}{3\sqrt{3}}$; in fact, the inequality is the other way round: $ABC\leq\frac{1}{3\sqrt{3}}$.
That's exactly what makes it such a tough problem. In fact, r can be 0, so it really doesn't help to know that $p^2\geq3$.

13. r can’t be 0 because a, b, and c (and hence A, B, and C) are all positive.

14. Originally Posted by JaneBennet
r can’t be 0 because a, b, and c (and hence A, B, and C) are all positive.
r can be arbitrarily close to zero (if for example a is very small, and b and c are close to 1/√3), so my previous comment still holds good.

15. By the way, some people think this problem is easy.

Art of Problem Solving Forum

Unfortunately I don’t understand the steps in their solution.

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