Inequality

**james_bond** · December 21st 2007, 07:49 AM

Prove that if $ab+bc+ca=\frac13$ ( $a,b,c$ are positive reals) then

$\frac a{a^2-bc+1}+\frac b{b^2-ca+1}+\frac c{c^2-ab+1}\ge \frac 1{a+b+c}$ .

**colby2152** · December 21st 2007, 09:25 AM

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

**JaneBennet** · December 22nd 2007, 05:41 AM

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

**james_bond** · December 22nd 2007, 08:33 AM

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.

**JaneBennet** · December 23rd 2007, 05:14 AM

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.

**Opalg** · December 23rd 2007, 12:48 PM

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?

**Isomorphism** · December 23rd 2007, 10:41 PM

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

**Opalg** · December 23rd 2007, 11:28 PM

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 $\frac{1}{3a+2}+\frac{1}{3b+2}+\frac{1}{3c+2}>1.03$ .

**Isomorphism** · December 23rd 2007, 11:39 PM

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

**Opalg** · December 29th 2007, 04:11 AM

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 $0<y\leqslant1$ (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.

**JaneBennet** · December 29th 2007, 07:25 AM

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: $ABC\leq\frac{1}{3\sqrt{3}}$ .

**Opalg** · December 29th 2007, 08:03 AM

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 $p^2\geq3$ .

**JaneBennet** · December 29th 2007, 10:45 AM

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

**Opalg** · December 29th 2007, 02:00 PM

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.

**JaneBennet** · December 29th 2007, 07:39 PM

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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