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Knowing that $a$, $b$ and $c$ are non-negatives, prove that

a/ $$(bc + ca + ab) \cdot \left[\frac{1}{(b - c)^2} + \frac{1}{(c - a)^2} + \frac{1}{(a - b)^2}\right] \ge 4$$

b/ $$(a^2 + b^2 + c^2) \cdot \left[\frac{1}{(b - c)^2} + \frac{1}{(c - a)^2} + \frac{1}{(a - b)^2}\right] \ge \frac{11 + 5 \sqrt 5}{2}$$

Well, it can be seen that $$[(a - b)^2 + (b - c)^2 + (c - a)^2] \cdot \left[\frac{1}{(b - c)^2} + \frac{1}{(c - a)^2} + \frac{1}{(a - b)^2}\right] \ge 3 + 5 \sqrt 5$$

But nothing more, because typically, $$(x + y + z) \cdot \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \ge 9, \forall x, y, z > 0$$

Other than that, I'm not sure.

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Lê Thành Đạt
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  • It should be $3+ 5\sqrt5$ instead of $19 + 5 \sqrt 5$ but it affects nothing – Taha Direk Nov 28 '19 at 15:39
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    2nd question answered here https://math.stackexchange.com/questions/2848219/find-min-of-p-dfrac1a-b2-dfrac1b-c2-dfrac1c-a2 – r9m Nov 28 '19 at 17:06
  • Unfortunately $\displaystyle \inf_{a,b,c \ge 0} ((a - b)^2 + (b - c)^2 + (c - a)^2) \left(\frac{1}{(b - c)^2} + \frac{1}{(c - a)^2} + \frac{1}{(a - b)^2}\right) = \frac{27}{2}$ .. however adding this with inequality (a) does not give you inequality (b). Inequality (b) is stronger in that sense. – r9m Nov 28 '19 at 22:37

1 Answers1

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One may consider the Buffalo Way because there are terms like $(a-b)$ which get simplified by the Buffalo substitution:

a) By cyclicity of the inequality we may assume $a\le b, c$. Hence, there are $x,y\geq 0$ such that $b=a+x$ and $c=a+y$. In fact, by well-definedness of your expression, we know that $x>0,y>0$ and $x\neq y$.

So your inequality is equivalent to $$\left(\frac{1}{x^2}+\frac{1}{(x-y)^2}+\frac{1}{y^2}\right) \big((a+x) (a+y)+a (a+x)+a (a+y)\big)-4\geq 0$$ which, by expansion and bringing everything on the same denominator, is the same as $$\frac{3 a^2 \left(x^2-x y+y^2\right)^2+2 a (x+y) \left(x^2-x y+y^2\right)^2+x y \left(x^2-3 x y+y^2\right)^2}{x^2 y^2 (x-y)^2}\geq 0.$$

The last inequality is trivially true because $a,x,y\geq 0$.

b) This one is more tricky. Using the same procedure as above we get that this inequality is equivalent to $$\frac1{2 x^2 y^2 (x-y)^2}\cdot X\geq0,$$ where $$X=6 a^2 \left(x^2-x y+y^2\right)^2+4 a (x+y) \left(x^2-x y+y^2\right)^2+2 x^6-4 x^5 y-\left(3+5 \sqrt{5}\right) x^4 y^2+2 \left(7+5 \sqrt{5}\right) x^3 y^3-\left(3+5 \sqrt{5}\right) x^2 y^4-4 x y^5+2 y^6.$$

So the original inequality is equivalent to $X\geq 0$. Note that \begin{split}X&\geq 2 x^6-4 x^5 y-\left(3+5 \sqrt{5}\right) x^4 y^2+2 \left(7+5 \sqrt{5}\right) x^3 y^3-\left(3+5 \sqrt{5}\right) x^2 y^4-4 x y^5+2 y^6\\ &=\left(x^2+\sqrt{5} \left(1+\frac{1}{\sqrt{5}}\right) x y+y^2\right) \left(x^2-\frac{1}{2} \sqrt{5} \left(1+\frac{3}{\sqrt{5}}\right) x y+y^2\right)^2 \\ &\geq0.\end{split}

As a conclusion: This is very far from a nice solution but it does the job.

Maximilian Janisch
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