# Long Inequality problem

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

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

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 $x+y+z\ge 6$

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

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.

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

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 $-1$.

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