6
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ What conditions on $a,b,c$? $\endgroup$ – Deepak Sep 29 '15 at 11:30
  • $\begingroup$ @Deepak for $a, b, c $ positive real numbers. I will edit the question $\endgroup$ – user260699 Sep 29 '15 at 11:33
  • $\begingroup$ Is it positive real number or positive integers? $\endgroup$ – tatan Sep 29 '15 at 16:12
  • $\begingroup$ @tatan postive real numbers $\endgroup$ – user260699 Sep 29 '15 at 18:29
1
$\begingroup$

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 $x+y+z\ge 6$

$\endgroup$
  • $\begingroup$ The last one is wrong, for example, $x=y =1, z = 4$. $\endgroup$ – River Li Oct 5 '19 at 2:58
  • $\begingroup$ You are right, thanks ! $\endgroup$ – Svetoslav Oct 5 '19 at 13:59
1
$\begingroup$

WLOG, assume that $a = \min(a,b,c)$.

After clearing the denominators, we need to prove that $$f(a, b, c) = Ac^2 + Bc + C\ge 0$$ where $A = a^2b + a(b-1)^2, \ B = (a-1)^2(b-1)^2 - a(a+b)$ and $C = a + b(a-1)^2$.

We split into two cases:

1) $a \ge \frac{3}{2}$: Let $a = \frac{3}{2} + x, \ b = \frac{3}{2} + y, \ c = \frac{3}{2} + z$ for $x, y, z \ge 0$. We have \begin{align} f(a, b, c) &= 32 x^2 y^2 z+32 x^2 y z^2+32 x y^2 z^2+48 x^2 y^2+128 x^2 y z +48 x^2 z^2\\ &\quad +128 x y^2 z+128 x y z^2+48 y^2 z^2+152 x^2 y + +120 x^2 z+120 x y^2\\ &\quad +384 x y z+152 x z^2+152 y^2 z+120 y z^2 + +120 x^2+320 x y+320 x z\\ &\quad +120 y^2+320 y z+120 z^2+218 x+218 y+218 z+117. \end{align} Clearly, we have $f(a,b,c)\ge 0$.

2) $a < \frac{3}{2}$: Note that $Ac^2 + Bc + C \ge 2\sqrt{AC}\, c + B c$. It suffices to prove that $2\sqrt{AC} + B \ge 0$ or $2\sqrt{AC} + (a-1)^2(b-1)^2 \ge a(a+b)$ or $$4 AC + (a-1)^4(b-1)^4 + 4(a-1)^2(b-1)^2 \sqrt{AC}\ge a^2(a+b)^2.$$ It suffices to prove that $$4 AC + (a-1)^4(b-1)^4 \ge a^2(a+b)^2. \tag{1}$$

There are two cases to deal with:

i) $b> 2$: With the substitutions $a = \frac{3}{2}\frac{u}{u+1}$ and $b = 2 + v$ for $u, v > 0$, (1) is written as $$\frac{1}{16(1+u)^4}(q_4v^4 + q_3v^3 + q_2v^2 + q_1v + q_0) \ge 0 \tag{2}$$ where \begin{align} q_4 &= (u-2)^4, \\ q_3 &= 4(7u^2+2u+4)(u-2)^2, \\ q_2 &= 246u^4-264u^3+396u^2+192u+96, \\ q_1 &= 520u^4-572u^3+816u^2+352u+64, \\ q_0 &= 328u^4-512u^3+600u^2+160u+16. \end{align} Clearly, $q_4, q_3 \ge 0$ for $u > 0$. Also, we have \begin{align} &q_2 > 246u^4-264u^3+396u^2 = 6u^2(41u^2-44u+66) \ge 0, \\ &q_1 > 520u^4-572u^3+816u^2 = 4u^2(130u^2-143u+204) \ge 0, \\ &q_0 > 328u^4-512u^3+600u^2 = 8u^2(41u^2-64u+75) \ge 0 \end{align} for $u > 0$. Thus, the inequality in (2) is true.

ii) $b\le 2$: With the substitution $a = \frac{3}{2}\frac{u}{u+1}$ for $u > 0$, (1) is written as $$\frac{1}{16(1+u)^4}(p_4u^4 + p_3u^3 + p_2u^2 + p_1u + p_0) \ge 0 \tag{3}$$ where \begin{align} p_4 &= b^4+20 b^3+102 b^2-160 b+64, \\ p_3 &= -8 b^4-40 b^3+168 b^2-508 b+280, \\ p_2 &= 24 b^4-96 b^3+396 b^2-384 b+168, \\ p_1 &= -32 (b^2-5 b+1) (b-1)^2, \\ p_0 &= 16 (b-1)^4. \end{align} Clearly, $p_0 \ge 0$. Also, we have \begin{align} &p_4 \ge 102 b^2-160 b+64 \ge 0, \\ &p_2 = 24b^2(b-2)^2 + 300 b^2-384 b+168 \ge 300 b^2-384 b+168 > 0. \end{align}

If $b \le \frac{3}{5}$, we have $$p_3 \ge -8 b^2-40 b^2+168 b^2-508 b+280 = 120b^2-508b+280 \ge 0, $$ and \begin{align} 4p_2p_0-p_1^2 &= 256(2b^4+16b^3-9b^2-56b+38)(b-1)^4 \\ &\ge 256(-9\cdot(\tfrac{3}{5})^2-56\cdot \tfrac{3}{5} +38)(b-1)^4\\ & \ge 0. \end{align} Thus, the inequality in (3) is true.

If $\frac{3}{5} < b \le 2$, we have $p_1 \ge 0$ and \begin{align} 4p_4p_2 - p_3^2 &= 16 (2 b^6+60 b^5+417 b^4-418 b^3+7167 b^2+492 b-2212) (b-1)^2\\ &\ge 16 (2 b^6+60 b^5+417 b^4-418 b^2\cdot 2+7167 b^2+492 b-2212) (b-1)^2\\ &= 16(2b^6+60b^5+417b^4+6331b^2+492b-2212)(b-1)^2\\ &\ge 16[2\cdot (\tfrac{3}{5})^6+60\cdot (\tfrac{3}{5})^5+417\cdot (\tfrac{3}{5})^4+6331 \cdot (\tfrac{3}{5})^2+492\cdot \tfrac{3}{5}-2212](b-1)^2\\ & \ge 0. \end{align} Thus, the inequality in (3) is true.

We are done.

$\endgroup$
  • $\begingroup$ Could you please explicitly demonstrate the validity of the inequalities $p_i$'s, $q_i$'s, (ii) and (iii)? $\endgroup$ – Hans Dec 18 '19 at 2:26
  • $\begingroup$ @Hans Let me update my answer. $\endgroup$ – River Li Dec 18 '19 at 2:31
  • $\begingroup$ @Hans I have updated my answer. Thanks for the comments. $\endgroup$ – River Li Dec 18 '19 at 6:02
  • $\begingroup$ +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? $\endgroup$ – Hans Dec 18 '19 at 23:51
  • $\begingroup$ @Hans I use Maple, Matlab and Mathematica. $\endgroup$ – River Li Dec 19 '19 at 1:33
-1
$\begingroup$

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 $-1$.

$\endgroup$
  • $\begingroup$ But this will give us the inequality is $>3$ not $ \geq 3$ $\endgroup$ – user260699 Sep 30 '15 at 17:28
  • $\begingroup$ @user260699-Don't know where I have got wrong...see if you can find any mistake.. $\endgroup$ – tatan Oct 1 '15 at 2:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.