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let $a,b,c>0$, show that $$\dfrac{a^2}{b(a^2-ab+b^2)}+\dfrac{b^2}{c(b^2-bc+c^2)}+\dfrac{c^2}{a(c^2-ca+a^2)}\ge\dfrac{9}{a+b+c}$$

My try: since this inequality is homogeneous ,without loss of generality, we assume that $$a+b+c=3$$ then

$$\Longleftrightarrow \sum_{cyc}\dfrac{a^2}{b(a^2-ab+b^2)}\ge 3$$

$$\Longleftrightarrow \sum_{cyc}\dfrac{a^2(a+b)}{b(a^3+b^3)}\ge 3$$

then I can't,Thank you

Michael Rozenberg
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  • I do not think you can assume that their sum is three, since that then limits the range of each of them to 0->3. – will Jan 16 '14 at 11:11
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    By the way, there's no problem to assume $a+b+c = 3$ at all. Homogenity argument allows one to assume $a+b+c = C$ for any positive C. –  Jan 19 '14 at 03:38
  • Setting $a=\frac1x$, and similarly $y,z$, the inequality becomes $$\sum_{cyc}\frac{y^3(x+y)}{x^3+y^3}\geq\frac{9xyz}{xy+yz+zx}.$$ This makes the LHS look nicer, but I didn't find a way out from here yet. Note that this is also equivalent to $$2(x+y+z)\geq\frac{9xyz}{xy+yz+zx}+\sum_{cyc}\frac{x^3(x+y)}{x^3+y^3}.$$ So finding an appropriate upperbound (if that's easier) is enough. – Bart Michels Jan 20 '14 at 12:48
  • This problem is routinely solvable using Lagrange multipliers. – J.R. Jan 25 '14 at 12:05

2 Answers2

5

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Proof without words ( partial answer / informal proof ) .
A formal proof has been given in another answer by another author.

  • Picture on the left: geometry of the conditions $a,b,c > 0$ and $a+b+c=3$ .
  • Picture in the middle: outside ($< 0$ : olive green) and inside ($\ge 0$ : white) of the function: $$f(a,b,c) = \sum_{cyc}\frac{a^2}{b(a^2-ab+b^2)} - \frac{9}{a+b+c}$$ as seen in the plane of the red triangle in the picture on the left.
    Since there are no green spots inside the triangle, the function is expected to be $\ge 0$ there.
  • Picture on the right: contour lines of $f(a,b,c)$ inside the triangle at levels $N = 1/2^k \;; \; k=0,\cdots,7$ . Contours are darker at lower level values.
    The minimum is expected to be $f(1,1,1)=0$ , which is at the center of the triangle.
Han de Bruijn
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3

Here is brutal force method which is ugly but solid.

the inequality equals:

$ \iff b^3c^6-ab^2c^6+a^2bc^6+a^3c^6-8ab^3c^5+8a^2b^2c^5-8a^3bc^5+8ab^4c^4+a^2b^3c^4-a^3b^2c^4+8a^4bc^4+b^6c^3-8ab^5c^3-a^2b^4c^3-6a^3b^3c^3+a^4b^2c^3-8a^5bc^3+a^6c^3+ab^6c^2+8a^2b^5c^2+a^3b^4c^2-a^4b^3c^2+8a^5b^2c^2-a^6bc^2-a^2b^6c-8a^3b^5c+8a^4b^4c-8a^5b^3c+a^6b^2c+a^3b^6+a^6b^3 \ge 0$

let $a=Min${$a,b,c$},$b=a+u,c=a+v,u \ge 0,v\ge0$ ,then we have:

$\iff 2(v^2-uv+u^2)a^7+2(2v^3-4uv^2+5u^2v+2u^3)a^6+6(v^4-3uv^3+2u^2v^2+3u^3v+u^4)a^5+(4v^5-9uv^4-8u^2v^3+26u^3v^2+19u^4v+4u^5)a^4+2(v^6-2uv^5+u^3v^3+13u^4v^2+4u^5v+u^6)a^3+2uv(v^5-2uv^4+4u^2v^3+3u^3v^2+4u^4v+2u^5)a^2+2u^2v^2(v^4-uv^3+4u^2v^2-u^3v+2u^4)a+u^3v^3(v^3+u^3) \ge 0$

the last one is true, and it is trivial when $u=v=0$, the "=" will hold.$\implies a=b=c$, the LHS=RHS.

QED.

chenbai
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