Prove that if $\displaystyle ab+bc+ca=\frac13$ ($\displaystyle a,b,c$ are positive reals) then
$\displaystyle \frac a{a^2-bc+1}+\frac b{b^2-ca+1}+\frac c{c^2-ab+1}\ge \frac 1{a+b+c}$.
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.
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:
$\displaystyle \sum{\frac{a^2+ab+ca}{a^2-bc+1}}\ \geq\ 1$
$\displaystyle \sum{\frac{a^2-bc+\frac{1}{3}}{a^2-bc+1}}\ \geq\ 1$
$\displaystyle \sum{\left[1-\frac{2}{3}\,\left(\frac{1}{a^2-bc+1}\right)\right]}\ \geq\ 1$
$\displaystyle \therefore\ \frac{1}{a^2-bc+1}+\frac{1}{b^2-ca+1}+\frac{1}{c^2-ab+1}\ \leq\ 3$
Now
$\displaystyle \frac{1}{a^2-bc+1}\ =\ \frac{1}{a^2+ab+ca+\frac{2}{3}}\ =\ \frac{3}{3a(a+b+c)+2}$
Thus
$\displaystyle \frac{1}{3a(a+b+c)+2}+\frac{1}{3b(a+b+c)+2}+\frac{ 1}{3c(a+b+c)+2}\ \leq\ 1$
Let $\displaystyle A=\sqrt{3}a,\ B=\sqrt{3}b,\ C=\sqrt{3}c$. Then $\displaystyle AB+BC+CA=1$ and the inequality becomes
$\displaystyle \color{white}.\hspace{15mm}\color{black}\frac{1}{A (A+B+C)+2}+\frac{1}{B(A+B+C)+2}+\frac{1}{C(A+B+C)+ 2}\ \leq\ 1$
This is what needs to be proved. I think the trick is to let $\displaystyle A=\tan{\alpha},\ B=\tan{\beta},\ C=\tan{\gamma}$, where $\displaystyle \alpha+\beta+\gamma=\frac{\pi}{2}$, but I still can’t do it.
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
$\displaystyle \color{white}.\hspace{15mm}\color{black}\frac{1}{A (A+B+C)+2} + \frac{1}{B(A+B+C)+2} + \frac{1}{C(A+B+C)+2}\ \leq\ 1$.
Let $\displaystyle x^3 - px^2 + qx -r = 0$ 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 $\displaystyle (x/p)^3 - p(x/p)^2 + (x/p) - r = 0$ has roots Ap, Bp and Cp. In other words, the equation $\displaystyle x^3 - p^2x^2 + p^2x - p^3r = 0$ has roots A(A+B+C), B(A+B+C), C(A+B+C).
Therefore the equation $\displaystyle (x-2)^3 - p^2(x-2)^2 + p^2(x-2) - p^3r = 0$ 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 $\displaystyle x^3 - (p^2+6)x^2 + (5p^2+12)x - (p^3r+6p^2+8) = 0$ has those roots.
Replacing x by 1/x, we see that the equation $\displaystyle (p^3r+6p^2+8)x^3 - (5p^2+12)x^2 + (p^2+6)x - 1 = 0$ has roots $\displaystyle \frac1{A(A+B+C)+2}$, $\displaystyle \frac1{B(A+B+C)+2}$, $\displaystyle \frac1{C(A+B+C)+2}$.
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 $\displaystyle \frac{5p^2+12}{p^3r+6p^2+8}\leqslant1$, where p=A+B+C and r=ABC. An equivalent inequality is $\displaystyle (rp+1)p^2\geqslant4$.
Now what?
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 $\displaystyle rp^3+p^2\geqslant4$. 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 $\displaystyle f(x,y,z) := xyz(x+y+z)^3 + (x+y+z)^2\geqslant4$. Let $\displaystyle D=\{(x,y,z)\in\mathbb{R}^3:x,y,z\geqslant0, yz+zx+xy=1\}$. 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 $\displaystyle (w+w^{-1})^2$, 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):
$\displaystyle \frac{\partial g}{\partial x} = yz(x+y+z)^3 + 3xyz(x+y+z)^2 +2(x+y+z) + \lambda(y+z) = 0$ ,
$\displaystyle \frac{\partial g}{\partial y} = zx(x+y+z)^3 + 3xyz(x+y+z)^2 +2(x+y+z) + \lambda(z+x) = 0$ ,
$\displaystyle \frac{\partial g}{\partial z} = xy(x+y+z)^3 + 3xyz(x+y+z)^2 +2(x+y+z) + \lambda(x+y) = 0$ .
Subtract the last of these from the middle one, to get $\displaystyle (z-y)\{x(x+y+z)^3 + \lambda\} = 0$ (and two similar equations with x,y and z cyclically interchanged). If x,y and z are all different then $\displaystyle \lambda = -x(x+y+z)^3 = -y(x+y+z)^3 = -z(x+y+z)^3$, 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 $\displaystyle x={\textstyle\frac12}(y^{-1}-y)$. Note that this implies $\displaystyle 0<y\leqslant1$ (otherwise x would be negative). Also, $\displaystyle x+y+y = \frac{1+3y^2}{2y}$ and $\displaystyle xy^2 = {\textstyle\frac12}y(1-y^2)$.
Then $\displaystyle f(x,y,y) = {\textstyle\frac12}y(1-y^2)\frac{(1+3y^2)^3}{(2y)^3} + \frac{(1+3y^2)^2}{(2y)^2}$, which simplifies(?) to $\displaystyle \frac{5+32y^2+54y^4-27y^8}{16y^2}$. We want this to be greater than or equal to 4, or in other words $\displaystyle 5+32y^2+54y^4-27y^8\geqslant64y^2$. When the term 64y^2 is taken over to the left side, this factorises as $\displaystyle (1-y^2)(1-3y^2)^2(5+3y^2)\geqslant0$, and this is clearly true when 0<y≤1.
Actually, it can be shown that $\displaystyle p^2\geq3$. By the Cauchy–Schwarz inequality,
$\displaystyle \sqrt{A^2+B^2+C^2}\sqrt{B^2+C^2+A^2}\ \geq\ AB+BC+CA\$
$\displaystyle \Rightarrow\ A^2+B^2+C^2 \geq\ 1$
$\displaystyle \Rightarrow\ (A+B+C)^2-2(AB+BC+CA)\ \geq\ 1$
$\displaystyle \Rightarrow\ (A+B+C)^2\ \geq\ 1+2\ =\ 3$
So it would appear to be simpler to prove that $\displaystyle rp^3\geq1$. The trouble is that $\displaystyle r\ngeq\frac{1}{3\sqrt{3}}$; in fact, the inequality is the other way round: $\displaystyle ABC\leq\frac{1}{3\sqrt{3}}$.
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.