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Source: old and nice inequality

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878 posts

#1 h May 19, 2019, 9:13 AM • 3 Y

Y by centslordm, Adventure10, Mango247

Let $a,b,c >0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+\frac{b}{c} +\frac{c}{a} \ge \frac{12}{1+3abc}$ Equality occurs when $a=b=c=1$ and also for $a \ge b \ge c$ are roots of the equation:
$x^3-3x^2+2x-\frac{1}{3}=0$ ,or any cyclic permutations thereof.

There is a proof of arqady here, with $uvw$ technique. Also, I found that this nice inequality can be proved easily

by SOS method, can anyone find it?

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878 posts

#2 h May 20, 2019, 5:34 PM • 1 Y

Y by Adventure10

2. Let $a,b,c$ be real number such that $a+b+c=3$ and $ab+bc+ca >0$ . Prove that:
$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} +\frac{9}{8(ab+bc+ca)} \ge \frac{15}{8}$ 3.(bel.jad5) Let $a,b,c \ge 0$ such that $a^2+b^2+c^2=3$ . Prove that:
$(a^3+3b^2c)^2+(b^3+3c^2a)^2+(c^3+3a^2b) ^2 \le 48$ 4.(Vasc) If $a,b,c$ are positive real numbers such that $a + b + c = 3$ , then
$\frac{24}{a^2b+b^2c+c^2a} +\frac{1}{abc} \ge 9$ Anyone's interested?

This post has been edited 3 times. Last edited by KaiRain, May 22, 2019, 3:32 PM

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#3 h May 22, 2019, 3:32 PM • 1 Y

Y by Adventure10

KaiRain wrote:

Let $a,b,c >0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+\frac{b}{c} +\frac{c}{a} \ge \frac{12}{1+3abc}$

We have: $162 \cdot [ (1+3abc)(ab^2+bc^2+ca^2)-12abc ]$

$= \sum_{cyc} (2c^2-ab+bc+7ca)(a^2-b^2-3ab+5bc-2ca)^2=S$

Note that: $\sum (a^2-b^2-3ab+5bc-2ca) = 0$ $\sum (2c^2-ab+bc+7ca) =2 \sum a^2+7 \sum bc >0$ $\sum (2c^2-\,ab\,+\,bc\,+\,7ca)(2a^2-\,bc\,+\,ca\,+\,7ab) = 3\sum a^2b^2\,+\, 12\sum ab^3\,+\, 66abc(a\,+\,b\,+\,c)>0$ Hence, $S \ge 0$ .

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#4 h May 23, 2019, 12:18 PM • 3 Y

Y by ali3985, Adventure10, Mango247

With computer, we obtain
$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} -\frac{12\left(\frac{a+b+c}{3}\right)^3}{3abc+\left(\frac{a+b+c}{3}\right)^3} = \frac{g(a,b,c)}{abc(a^3+3a^2b+3a^2c+3ab^2+87abc+3ac^2+b^3+3b^2c+3bc^2+c^3)}$ where
$g(a,b,c) = a b (2 a b-5 a c-b^2+3 b c+c^2)^2 + bc (a^2-5 a b+3 a c+2 b c-c^2)^2+c a (a^2-3 a b-2 a c-b^2+5 b c)^2$ $+\frac{1}{4} (2 a^2 b-4 a^2 c-a b^2-a c^2-b^2 c+5 b c^2)^2+\frac{3}{4} (2 a^2 c-3 a b^2-a c^2+b^2 c+b c^2)^2.$

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#5 h May 23, 2019, 2:28 PM • 3 Y

Y by KaiRain, Adventure10, Mango247

(Vasc) If $a, b, c$ are positive real numbers such that $a+b+c=3$ , then $\frac{24}{a^2b+b^2c+c^2a} + \frac{1}{abc} \ge 9$ .

Proof: The method was used in my old post: https://artofproblemsolving.com/community/u473756h1842210p12397448

Let $p = a+b+c, \ q = ab+bc+ca$ and $r = abc$ .
Let $\Delta = -4p^3r+p^2q^2+18pqr-4q^3-27r^2$ . It is easy to verify that $\Delta = (a-b)^2(b-c)^2(c-a)^2$ . Thus, we have $\Delta \ge 0$ .

Let $x = a^2b+b^2c+c^2a$ .
It is easy to verify that $x^2+(-pq+3r)x+p^3r-6pqr+q^3+9r^2 = 0$ . This quadratic equation about $x$ has two real roots $x_{1, 2} = \frac{pq-3r\pm \sqrt{\Delta}}{2}.$
Thus, we have $x\le \frac{pq-3r+\sqrt{\Delta}}{2}.$ There are two possible cases:

(1) $r \le \frac{1}{9}$ : The inequality is true clearly.

(2) $r > \frac{1}{9}$ : It suffices to prove that $\frac{pq-3r+\sqrt{\Delta}}{2} \le \frac{24r}{9r - 1}.$ It suffices to prove the following two inequalities:
i) $\frac{48r}{9r - 1} - pq + 3r \ge 0$ ;
and ii) $(\frac{48r}{9r - 1} - pq + 3r)^2 \ge \Delta.$

i) is equivalent to $\frac{3r(3r+5)}{9r-1}\ge q$ . According to Schur's inequality, we have $p^3+9r\ge 4pq$ or $q \le \frac{3r+9}{4}.$
It suffices to prove that $\frac{3r(3r+5)}{9r-1} \ge \frac{3r+9}{4}$ or $\frac{9(r-1)^2}{4(9r-1)}\ge 0$ . It is true.

ii) is equivalent to
$(9r-1)^2q^3 - 54r(3r+1)(9r-1)q + 729r^4+2673r^3+27r^2+27r \ge 0.$ Let $A = (9r-1)^2,\ B = 54r(3r+1)(9r-1), \ C = 729r^4+2673r^3+27r^2+27r.$
It suffices to prove that $Aq^3 + \frac{C}{2} + \frac{C}{2} \ge 3\sqrt[3]{Aq^3\frac{C}{2}\frac{C}{2}} \ge Bq.$
It suffices to prove that $27AC^2 - 4B^3\ge 0$ or $19683r^2(r-1)^2(9r-1)^2(27r^2-18r-1)^2\ge 0.$ We are done.

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878 posts

#6 h May 25, 2019, 5:12 AM • 2 Y

Y by Adventure10, Mango247

dragonheart6 wrote:

$g(a,b,c) = a b (2 a b-5 a c-b^2+3 b c+c^2)^2 + bc (a^2-5 a b+3 a c+2 b c-c^2)^2+c a (a^2-3 a b-2 a c-b^2+5 b c)^2$ $+\frac{1}{4} (2 a^2 b-4 a^2 c-a b^2-a c^2-b^2 c+5 b c^2)^2+\frac{3}{4} (2 a^2 c-3 a b^2-a c^2+b^2 c+b c^2)^2.$

Your indentity is impressive, dragonheart6. Can you write 2, 3 and 4 in the form Sum of square like that?

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1442 posts

#7 h May 25, 2019, 6:31 AM • 1 Y

Y by Adventure10

KaiRain wrote:

dragonheart6 wrote:

$g(a,b,c) = a b (2 a b-5 a c-b^2+3 b c+c^2)^2 + bc (a^2-5 a b+3 a c+2 b c-c^2)^2+c a (a^2-3 a b-2 a c-b^2+5 b c)^2$ $+\frac{1}{4} (2 a^2 b-4 a^2 c-a b^2-a c^2-b^2 c+5 b c^2)^2+\frac{3}{4} (2 a^2 c-3 a b^2-a c^2+b^2 c+b c^2)^2.$

Your indentity is impressive, dragonheart6. Can you write 2, 3 and 4 in the form Sum of square like that?

Usually not easy. P2 can be proved by the SHED technique (Buffalo Way).

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878 posts

#8 h May 25, 2019, 12:55 PM • 1 Y

Y by Adventure10

I can prove 2 and 3 in the similar way (see the links above). But for 4, it's very ugly

:blush:

This post has been edited 1 time. Last edited by KaiRain, May 25, 2019, 1:07 PM

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#9 h May 25, 2019, 1:04 PM • 2 Y

Y by Adventure10, Mango247

Can anyone find a nice identity in form SOS for this old one ?
5.(canhang_2007) Let $a,$ $b,$ $c$ be positive real numbers such that $a+b+c=1$ . Prove that
$\frac{36}{a^2b+b^2c+c^2a}+\frac{1}{abc} \ge 343$

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#11 h May 25, 2019, 1:17 PM • 1 Y

Y by Adventure10

Dear Nguyenhuyen_AG and xzlbq, can you try it?

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#12 h May 25, 2019, 3:14 PM • 1 Y

Y by Adventure10

KaiRain wrote:

I can prove 2 and 3 in the similar way (see the links above). But for 4, it's very ugly

:blush:

P3. $\frac{16}{9}(a^2+b^2+c^2)^3-(a^3+3b^2c)^2-(3ac^2+b^3)^2-(3a^2b+c^3)^2$ $= \frac{1}{9}(a^3-a^2 b-2 a^2 c-2 a b^2+6 a b c-a c^2+b^3-b^2 c-2 b c^2+c^3)^2+\frac{1}{18} (3 a^3+a^2 b-8 a b^2-2 a c^2-2 b^3+b^2 c+8 b c^2-c^3)^2$ $+\frac{1}{9}(a^2 b-7 a^2 c+3 a b^2-2 b^3-b^2 c+4 b c^2+2 c^3)^2+\frac{1}{18} (c+b+a)^2 (a^2-2 a b+a c+b c-c^2)^2+\frac{1}{9}(a^3+a^2 c-3 a b^2+2 b c^2-c^3)^2.$

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878 posts

#13 h Sep 15, 2019, 3:28 AM • 1 Y

Y by Adventure10

6. Let $a,b,c>0$ , prove that:
$\frac{(a+2b)(b+2c)(c+2a)}{abc}+\frac{1}{8} \ge \frac{637}{8} \cdot \frac{a^2+b^2+c^2}{(a+b+c)^2}$

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878 posts

#14 h Sep 15, 2019, 3:31 AM • 1 Y

Y by Adventure10

7.(VQBC) Let $a,b,c$ be non-negative numbers such that $a+b+c=3$ . Prove that:
$a^2b+b^2c+c^2a+abc \cdot [1+\frac{1}{6}(a^2+b^2+c^2-ab-bc-ca)] \le 4$

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3316 posts

#15 h Sep 15, 2019, 4:00 AM • 3 Y

Y by KaiRain, Adventure10, Mango247

KaiRain wrote:

7.(VQBC) Let $a,b,c$ be non-negative numbers such that $a+b+c=3$ . Prove that:
$a^2b+b^2c+c^2a+abc \cdot [1+\frac{1}{6}(a^2+b^2+c^2-ab-bc-ca)] \le 4$

https://artofproblemsolving.com/community/c6h403239
There are AM-GM proof but don't belong me.

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878 posts

#16 h Sep 15, 2019, 7:31 AM • 2 Y

Y by dragonheart6, Adventure10

KaiRain wrote:

7.(VQBC) Let $a,b,c$ be non-negative numbers such that $a+b+c=3$ . Prove that:
$a^2b+b^2c+c^2a+abc \cdot [1+\frac{1}{6}(a^2+b^2+c^2-ab-bc-ca)] \le 4$

We have:
$8(a+b+c)^5-54(a+b+c)^2(a^2b+b^2c+c^2a+abc)-81abc(a^2+b^2+c^2-ab-bc-ca)$
$=\frac{1}{2} \sum_{cyc}(5b+5c-a)(2a^2-2b^2-3ab+4bc-ca)^2 \ge 0$
Click to reveal hidden text

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878 posts

#17 h Sep 15, 2019, 7:36 AM • 1 Y

Y by Adventure10

Nguyenhuyen_AG wrote:

https://artofproblemsolving.com/community/c6h403239
There are AM-GM proof but don't belong me.

Thank you Nguyenhuyen_AG, do you have any link about it?

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780 posts

SBM

#18 h Jun 25, 2020, 3:28 AM • 1 Y

Y by KaiRain

KaiRain wrote:

KaiRain wrote:

7.(VQBC) Let $a,b,c$ be non-negative numbers such that $a+b+c=3$ . Prove that:
$a^2b+b^2c+c^2a+abc \cdot [1+\frac{1}{6}(a^2+b^2+c^2-ab-bc-ca)] \le 4$

We have:
$8(a+b+c)^5-54(a+b+c)^2(a^2b+b^2c+c^2a+abc)-81abc(a^2+b^2+c^2-ab-bc-ca)$
$=\frac{1}{2} \sum_{cyc}(5b+5c-a)(2a^2-2b^2-3ab+4bc-ca)^2 \ge 0$
Click to reveal hidden text

Better SOS is $:$

$8(a+b+c)^5-54(a+b+c)^2(a^2b+b^2c+c^2a+abc)-81abc(a^2+b^2+c^2-ab-bc-ca)$ $=\frac{1}{6} \sum\limits_{cyc} \left( a+b+7\,c \right) \left( a+b-2\,c \right) ^{2} \left( 2\, a-4\,b-c \right) ^{2}$
How do you find $[2a^2-2b^2-3ab+4bc-ca,...],$ KaiRain $?$

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3316 posts

#20 h Jun 25, 2020, 4:56 AM • 1 Y

Y by dragonheart6

Another result from KaiRain skill
$8(a+b+c)^5-54(a+b+c)^2(a^2b+b^2c+c^2a+abc)-81abc(a^2+b^2+c^2-ab-bc-ca)$ $= \frac{1}{338} \sum (83a+41b-7c)(6a^2+2b^2-8c^2-13ab+13ca)^2.$

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SBM

#22 h Jun 25, 2020, 6:32 AM • 1 Y

Y by dragonheart6

Nguyenhuyen_AG wrote:

Another result from KaiRain skill
$\text{P}=8(a+b+c)^5-54(a+b+c)^2(a^2b+b^2c+c^2a+abc)-81abc(a^2+b^2+c^2-ab-bc-ca)$ $= \frac{1}{338} \sum (83a+41b-7c)(6a^2+2b^2-8c^2-13ab+13ca)^2.$

There are too much similar KaiRain's form. For example $:$
$\begin{align} \text{P}&=\sum {\frac { \left( 43\,a-5\,b+25\,c \right) \left( 2\,{a}^{2}-8\,ab+19\, ca+8\,{b}^{2}-11\,bc-10\,{c}^{2} \right) ^{2}}{294}} \\& =\sum {\frac { \left( 85\,a+37\,b-5\,c \right) \left( 10\,{a}^{2}-22\,ab+23 \,ca+4\,{b}^{2}-bc-14\,{c}^{2} \right) ^{2}}{1014}} \\&=\sum {\frac { \left( 131\,a+5\,b+35\,c \right) \left( 4\,{a}^{2}-11\,ab+18 \,ca+6\,{b}^{2}-7\,bc-10\,{c}^{2} \right) ^{2}}{722}} \\&=\sum {\frac { \left( 331\,a+277\,b-59\,c \right) \left( 26\,{a}^{2}-53\,ab +43\,ca+2\,{b}^{2}+10\,bc-28\,{c}^{2} \right) ^{2}}{22326}} \\&=\sum {\frac { \left( 85\,a+37\,b-5\,c \right) \left( 10\,{a}^{2}-22\,ab+23 \,ca+4\,{b}^{2}-bc-14\,{c}^{2} \right) ^{2}}{1014}} \\&=\sum {\frac { \left( 335\,a+5\,b+101\,c \right) \left( 6\,{a}^{2}-17\,ab+ 29\,ca+10\,{b}^{2}-12\,bc-16\,{c}^{2} \right) ^{2}}{4802}} \\&=\sum {\frac { \left( 4529\,a-865\,b+4175\,c \right) \left( 2\,{a}^{2}-33\, ab+119\,ca+58\,{b}^{2}-86\,bc-60\,{c}^{2} \right) ^{2}}{1517282}} \\&=\sum {\frac { \left( 521\,a-25\,b+215\,c \right) \left( 6\,{a}^{2}-19\,ab+ 37\,ca+14\,{b}^{2}-18\,bc-20\,{c}^{2} \right) ^{2}}{12482}} \\&=\sum {\frac { \left( 18185\,a-301\,b+6425\,c \right) \left( 40\,{a}^{2}- 119\,ab+216\,ca+78\,{b}^{2}-97\,bc-118\,{c}^{2} \right) ^{2}}{14590802 }} \\&=\sum {\frac { \left( 2773\,a-185\,b+1255\,c \right) \left( 22\,{a}^{2}-73 \,ab+149\,ca+58\,{b}^{2}-76\,bc-80\,{c}^{2} \right) ^{2}}{1093974}} \\&=\sum {\frac { \left( 83\,a-7\,b+41\,c \right) \left( 2\,{a}^{2}-7\,ab+15\, ca+6\,{b}^{2}-8\,bc-8\,{c}^{2} \right) ^{2}}{338}} \\&=\sum {\frac { \left( 425\,a+35\,b+89\,c \right) \left( 8\,{a}^{2}-21\,ab+ 32\,ca+10\,{b}^{2}-11\,bc-18\,{c}^{2} \right) ^{2}}{7442}} \end{align}$

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#24 h Jun 29, 2020, 6:21 PM • 2 Y

Y by dragonheart6, Mango247

dragonheart6 wrote:

P3. $\frac{16}{9}(a^2+b^2+c^2)^3-(a^3+3b^2c)^2-(3ac^2+b^3)^2-(3a^2b+c^3)^2$ $= \frac{1}{9}(a^3-a^2 b-2 a^2 c-2 a b^2+6 a b c-a c^2+b^3-b^2 c-2 b c^2+c^3)^2+\frac{1}{18} (3 a^3+a^2 b-8 a b^2-2 a c^2-2 b^3+b^2 c+8 b c^2-c^3)^2$ $+\frac{1}{9}(a^2 b-7 a^2 c+3 a b^2-2 b^3-b^2 c+4 b c^2+2 c^3)^2+\frac{1}{18} (c+b+a)^2 (a^2-2 a b+a c+b c-c^2)^2+\frac{1}{9}(a^3+a^2 c-3 a b^2+2 b c^2-c^3)^2.$

Another SOS's proof is:
$(\sum_{\text{cyc}} a^2-\sum_{\text{cyc}} bc) \cdot [16 \sum_{\text{cyc}} a^2 - 9 \sum_{\text{cyc}} (a^3+3b^2c)^2]$ $=\frac{1}{45} \ (\sum_{\text{cyc}} a^2-\sum_{\text{cyc}} bc) \cdot (\sum_{\text{cyc}} a^3-5\sum_{\text{cyc}} a^2b+4\sum_{\text{cyc}}ab^2)^2$ $+\ \frac{1}{90} \ ( 5\sum_{\text{cyc}} a^4\,-\,32\sum_{\text{cyc}}a^3b\,+\,85\sum_{\text{cyc}} ab^3\,-\,41\sum_{\text{cyc}} a^2b^2\ -\ 17abc\sum_{\text{cyc}} a)^2$ $+\ \frac{1}{10}\ [16\sum_{\text{cyc}} a^2\ +\ 51(\sum_{\text{cyc}} a)^2] \ (\sum_{\text{cyc}} a^3\ -\ \sum_{\text{cyc}} a^2b\ -\ 2\sum_{\text{cyc}}ab^2\ +\ 6abc)^2 \ \ge 0.$

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SBM

#27 h Mar 7, 2021, 2:16 AM

Y by

dragonheart6 wrote:

With computer, we obtain
$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} -\frac{12\left(\frac{a+b+c}{3}\right)^3}{3abc+\left(\frac{a+b+c}{3}\right)^3} = \frac{g(a,b,c)}{abc(a^3+3a^2b+3a^2c+3ab^2+87abc+3ac^2+b^3+3b^2c+3bc^2+c^3)}$ where
$g(a,b,c) = a b (2 a b-5 a c-b^2+3 b c+c^2)^2 + bc (a^2-5 a b+3 a c+2 b c-c^2)^2+c a (a^2-3 a b-2 a c-b^2+5 b c)^2$ $+\frac{1}{4} (2 a^2 b-4 a^2 c-a b^2-a c^2-b^2 c+5 b c^2)^2+\frac{3}{4} (2 a^2 c-3 a b^2-a c^2+b^2 c+b c^2)^2.$

How how how?

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1442 posts

#28 h Mar 7, 2021, 2:56 AM • 3 Y

Y by Mango247, Mango247, Mango247

SBM wrote:

dragonheart6 wrote:

With computer, we obtain
$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} -\frac{12\left(\frac{a+b+c}{3}\right)^3}{3abc+\left(\frac{a+b+c}{3}\right)^3} = \frac{g(a,b,c)}{abc(a^3+3a^2b+3a^2c+3ab^2+87abc+3ac^2+b^3+3b^2c+3bc^2+c^3)}$ where
$g(a,b,c) = a b (2 a b-5 a c-b^2+3 b c+c^2)^2 + bc (a^2-5 a b+3 a c+2 b c-c^2)^2+c a (a^2-3 a b-2 a c-b^2+5 b c)^2$ $+\frac{1}{4} (2 a^2 b-4 a^2 c-a b^2-a c^2-b^2 c+5 b c^2)^2+\frac{3}{4} (2 a^2 c-3 a b^2-a c^2+b^2 c+b c^2)^2.$

How how how?

$(a, b, c) \to (a^2, b^2, c^2)$ ,

option.solver = 'sedumi';
[Q, Z] = findsos(p,'rational',option);

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878 posts

#30 h Jun 13, 2021, 7:17 AM • 4 Y

Y by Nguyenhuyen_AG, dragonheart6, ali3985, jokehim

KaiRain wrote:

Let $a,b,c >0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+\frac{b}{c} +\frac{c}{a} \ge \frac{12}{1+3abc}$

Today I found a nice proof, hope you like it.
Using C-S:
$\begin{align*} \frac{a}{b} +\frac{b}{c}+\frac{c}{a} \ge \frac{ \left[ \sum a \left(2a-b+5c \right) \right]^2}{ \sum ab \left(2a-b+5c \right)^2}= \frac{4 \left(a+b+c \right)^4}{\sum ab \left(2a-b+5c \right)^2},\end{align*}$ it remains to prove a cyclic inequality of degree 4, which is:
$\begin{align*} \left(a+b+c \right)^4 +81 abc \left(a+b+c \right) \ge \sum_{\text{cyc}} ab \left(2a-b+5c \right)^2 , \end{align*}$ or $\begin{align*} \sum_{\text{cyc}} \left(a^4-8a^3b+18a^2b^2-12a^2bc +ab^3\right) \ge 0, \end{align*}$ $\begin{align*} \sum_{\text{cyc}} \left(a^2-b^2-3ab+5bc-2ca \right)^2 \ge 0. \end{align*}$ This end the proof.

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#31 h Jun 13, 2021, 3:09 PM

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KaiRain wrote:

Cauchy-Schwarz proof

I like it

https://voz.vn/styles/next/xenforo/smilies/popopo/beauty.png

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1442 posts

#32 h Jun 14, 2021, 8:42 AM

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KaiRain wrote:

KaiRain wrote:

Let $a,b,c >0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+\frac{b}{c} +\frac{c}{a} \ge \frac{12}{1+3abc}$

Today I found a nice proof, hope you like it.
Using C-S:
Click to reveal hidden text

Very nice. Can this approach be used for some inequalities involving $a/b + b/c + c/a$ e.g. in AoPS by designing appropriate C-S terms?

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arqady

#33 h Jun 14, 2021, 11:16 AM

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No. I checked it.
For example, for this inequality we obtain:
$\sum_{cyc}\frac{a^2}{b}=\sum_{cyc}\frac{a^2(2a-b+2c)^2}{b(2a-b+2c)^2}\geq\frac{\left(\sum\limits_{cyc}(2a^2+ab)\right)^2}{\sum\limits_{cyc}b(2a-b+2c)^2},$ which saves the equality occurring case, but gives a harder inequality.

This post has been edited 3 times. Last edited by arqady, Jun 14, 2021, 11:26 AM

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#34 h Jun 14, 2021, 12:08 PM

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I agree, sometimes it leads to high degree.

arqady wrote:

For example, for this inequality we obtain:
$\sum_{cyc}\frac{a^2}{b}=\sum_{cyc}\frac{a^2(2a-b+2c)^2}{b(2a-b+2c)^2}\geq\frac{\left(\sum\limits_{cyc}(2a^2+ab)\right)^2}{\sum\limits_{cyc}b(2a-b+2c)^2},$ which saves the equality occurring case, but gives a harder inequality.

Did you have a try with $\sum_{cyc} \frac{\left(a+ \frac{b}{2} \right)^2}{b},$ there is something beautiful

.

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#35 h Jun 14, 2021, 1:00 PM

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KaiRain wrote:

KaiRain wrote:

Let $a,b,c >0$ such that $a+b+c=3$ . Prove that:
$\frac{a}{b}+\frac{b}{c} +\frac{c}{a} \ge \frac{12}{1+3abc}$

Today I found a nice proof, hope you like it.
Using C-S:
$\begin{align*} \frac{a}{b} +\frac{b}{c}+\frac{c}{a} \ge \frac{ \left[ \sum a \left(2a-b+5c \right) \right]^2}{ \sum ab \left(2a-b+5c \right)^2}= \frac{4 \left(a+b+c \right)^4}{\sum ab \left(2a-b+5c \right)^2},\end{align*}$ it remains to prove a cyclic inequality of degree 4, which is:
$\begin{align*} \left(a+b+c \right)^4 +81 abc \left(a+b+c \right) \ge \sum_{\text{cyc}} ab \left(2a-b+5c \right)^2 , \end{align*}$ or $\begin{align*} \sum_{\text{cyc}} \left(a^4-8a^3b+18a^2b^2-12a^2bc +ab^3\right) \ge 0, \end{align*}$ $\begin{align*} \sum_{\text{cyc}} \left(a^2-b^2-3ab+5bc-2ca \right)^2 \ge 0. \end{align*}$ This end the proof.

:first:

This post has been edited 1 time. Last edited by ali3985, Jun 14, 2021, 1:02 PM

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