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for $a, b, c $ positive real numbers $$ \left( a+\frac{1}{b} -1\right) \left( b+\frac{1}{c} - 1\right) +\left( b+\frac{1}{c} -1\right) \left( c+\frac{1}{a} -1\right)$$ $$ +\left( c+\frac{1}{a} -1\right) \left( a+\frac{1}{b} -1\right) \geq 3$$

How we can prove the inequality above. Actually it take long time to prove it but I couldn't complete. How we prove it? . Thanks for help

River Li
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user260699
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3 Answers3

1

Maybe this could help:

If you denote $x=a+\frac{1}{b}>0, \, y=b+\frac{1}{c}>0, \, z=c+\frac{1}{a}>0$ you will have $x+y+z=a+1/a +b+1/b+c+1/c\ge 2+2+2=6$ and the equality is achieved iff $a=b=c=1$

You get $(x-1)(y-1)+(y-1)(z-1)+(z-1)(x-1)\ge 3$ which is equivalent to $$xy+yz+zx-2(x+y+z)\ge 0$$ So you have to prove the last one, having in mind that $x,y,z>0$ and $x+y+z\ge 6$

Svetoslav
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1

Edit: 2023/06/17 We give a new proof.

Let $$x = a + 1/b, \quad y = b + 1/c, \quad z = c + 1/a.$$

It suffices to prove that $$(x - 1)(y-1) + (y-1)(z-1) + (z-1)(x - 1) \ge 3,$$ or $$xy + yz + zx - 2(x + y + z) \ge 0. \tag{1} $$

To this end, note that (see Remarks at the end) $$xyz - x - y - z - 2 = \frac{(abc - 1)^2}{abc} \ge 0. \tag{2}$$

Fact 1: Let $u, v, w \ge 0$ with $uvw - u - v - w - 2 \ge 0$. Then $uv + vw + wu - 2(u + v + w) \ge 0.$ (The proof is given at the end.)

Using Fact 1, letting $u = x, v = y, w = z$, we know that (1) is true.

We are done.

$\phantom{2}$

Remarks:

How do we find $xyz - x - y - z - 2$?

From $x = a + 1/b, y = b + 1/c, z = c + 1/a$, after eliminating $a, b$, we have $$\left( xy-1 \right) {c}^{2}+ \left( -xyz-x+y+z \right) c+xz-1 = 0.$$ It is a quadratic equation in $c$ with the discriminant $$\Delta = (xyz - x - y - z + 2)(xyz - x - y - z - 2).$$ It must hold that $\Delta \ge 0$. Then I calculate $xyz - x - y - z - 2 = \frac{(abc - 1)^2}{abc} \ge 0$.

I notice that if $x, y, z \ge 0$ with $xyz - x - y - z - 2 \ge 0$, then $xy + yz + zx - 2(x + y + z) \ge 0$.

$\phantom{2}$

Proof of Fact 1:

From $uvw - u - v - w - 2 \ge 0$, we have $$(uv - 1)w \ge u + v + 2$$ which results in $$uv > 1, \quad w \ge \frac{u + v + 2}{uv - 1}. \tag{3}$$

The desired inequality is written as $$(u + v - 2)w \ge 2u + 2v - uv. \tag{4}$$

Using (3), we have $u + v \ge 2\sqrt{uv} > 2$. Using (3) and (4), it suffices to prove that $$(u + v - 2)\cdot \frac{u + v + 2}{uv - 1} \ge 2u + 2v - uv $$ or $$\frac{(u + v + 1 - uv)^2 + uv - 5}{uv - 1} \ge 0. \tag{5}$$

If $uv \ge 5$, clearly (5) is true.

If $uv < 5$, we have $$u + v + 1 - uv \ge 2\sqrt{uv} + 1 - uv = 2 - (\sqrt{uv} - 1)^2 > 0$$ and thus \begin{align*} (u + v + 1 - uv)^2 + uv - 5 &\ge (2\sqrt{uv} + 1 - uv)^2 + uv - 5\\ &= (uv - 1)(\sqrt{uv} - 2)^2 \\ &\ge 0. \end{align*} Thus, (5) is true.

We are done.

River Li
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  • Could you please explicitly demonstrate the validity of the inequalities $p_i$'s, $q_i$'s, (ii) and (iii)? – Hans Dec 18 '19 at 02:26
  • @Hans Let me update my answer. – River Li Dec 18 '19 at 02:31
  • @Hans I have updated my answer. Thanks for the comments. – River Li Dec 18 '19 at 06:02
  • +1, impressive! It must have been a lot of trial and error to sift through all the cases. Do you have a system to do so? – Hans Dec 18 '19 at 23:51
  • @Hans I use Maple, Matlab and Mathematica. – River Li Dec 19 '19 at 01:33
  • Would you terribly mind putting in some comments at different stages, especially the critical points of $a$ at $\frac32$ and $b$ at $\frac35$ and $2$, of the inequalities to point out how you use the various tools that you mentioned you used? It would really help me understand your motivation for the case splitting and learn the technique. I would truly appreciate it! – Hans Dec 19 '19 at 02:03
  • @Hans Simply use Matlab to do simulation. For example, for $a < ?$, (1) holds. Simulation shows $a < \frac{3}{2}$ works (Other value rather than $\frac{3}{2}$ also works, but $\frac{3}{2}$ is simpler) . For $b$, first I want to find $b$ such that the coefficients in (3) are all non-negative, but fail. Then I want to find $b$ such that $p_3\ge 0$ and $4p_2p_0\ge p_1^2$, or $p_1\ge 0$ and $4p_4p_2\ge p_3^2$. – River Li Dec 19 '19 at 02:14
  • I see. Thank you. – Hans Dec 19 '19 at 17:53
  • @Hans My old proof is very complicated. I give a simple proof just now. – River Li Jun 17 '23 at 05:12
  • Very crafty. I would upvote again had I one more point. I wonder if the reason you used the expression $xyz-x-y-z-2$ for the constraint by looking at $xyz$ because of the symmetry. – Hans Jun 18 '23 at 04:56
  • @Hans Nice point. I will add something about $xyz - x - y - z - 2$. – River Li Jun 18 '23 at 05:41
  • Very nice. Thank you. Previously I got to the part expressing $c$ in terms of $(x,y,z)$ after eliminating $(a,b)$ but did not reach the discriminant and come up with $xyz - x - y - z - 2$. – Hans Jun 20 '23 at 04:23
  • @Hans Actually, I found it by chance. For me, it is surprising that $x, y, z\ge 0$ and $xyz - x - y - z - 2$ imply $xy + yz + zx - 2(x + y + z) \ge 0$, because $xyz - x - y - z - 2 \ge 0$ is merely the requirement of $\Delta \ge 0$ in the system $x = a + 1/b, y = b + 1/c, z = c + 1/a$. Thus, there is a question: If $x, y, z\ge 0$ and $xyz - x - y - z -2 \ge 0$, are there $a, b, c > 0$ such that $x = a + 1/b, y = b + 1/c, z = c + 1/a$? – River Li Jun 20 '23 at 05:47
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    I see. Yes, it is surprising that the necessary condition suffices to prove the desired inequality. Maybe you could pose this existence problem as a math.exchange question. Let me know the link to the question when you do it. – Hans Jun 25 '23 at 16:59
-1

Hint:

$a,b,c$ are positive real numbers.

So,$a+\frac 1b>0$

So,$a+\frac 1b-1>-1$.

Similarly all the values are greater than $-1$.

Soham
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