Inequality with 3 variables

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Inequality with 3 variables

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Source: alijadallah belabess

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3750 posts

#1 h Nov 4, 2018, 3:03 PM • 2 Y

Y by nguyendangkhoa17112003, Adventure10

Let $a$ , $b$ and $c$ positive real numbers such that: $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Prove that:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3 \geq 2(a+b+c)$

This post has been edited 1 time. Last edited by bel.jad5, Nov 4, 2018, 3:03 PM

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#2 h Nov 4, 2018, 4:14 PM • 3 Y

Y by ywq233, Adventure10, Mango247

[(ab(a+c)+bc(b+a)+ca(c+b)]^2>=4(ab+bc+ca)[(ab)(bc)+(bc)(ca)+(ca)(ab)]

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3750 posts

#3 h Nov 10, 2018, 5:47 PM • 2 Y

Y by Adventure10, Mango247

luofangxiang wrote:

[(ab(a+c)+bc(b+a)+ca(c+b)]^2>=4(ab+bc+ca)[(ab)(bc)+(bc)(ca)+(ca)(ab)]

And?

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4613 posts

#4 h Nov 10, 2018, 5:59 PM • 2 Y

Y by ywq233, Adventure10

[a(x+y)+b(y+z)+c(z+x)]^2>=4(ab+bc+ca)(xy+yz+zx)

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#5 h Nov 10, 2018, 11:34 PM • 3 Y

Y by bel.jad5, luofangxiang, Adventure10

bel.jad5 wrote:

Let $a$ , $b$ and $c$ positive real numbers such that: $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Prove that:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3 \geq 2(a+b+c)$

Use $AX-BY\ge \sqrt{A^2-B^2}.\sqrt{X^2-Y^2},(*),(A\ge B \ge 0, X\ge Y\ge 0)$ Because $(*) \Leftrightarrow (AY-BX)^2\ge 0 , (true)$ $LHS=\frac{ab^2+bc^2+ca^2+3abc}{abc}=\frac{(a+b+c)(ab+bc+ca)-(a^2b+b^2c+c^2a)}{abc}$ $LHS \ge \frac{(a+b+c)(ab+bc+ca)-\sqrt{a^2+b^2+c^2}.\sqrt{a^2b^2+b^2c^2+c^2a^2}}{abc}$ $LHS \ge \frac{\sqrt{(a+b+c)^2-(\sqrt{a^2+b^2+c^2})^2}.\sqrt{(ab+bc+ca)^2-(\sqrt{a^2b^2+b^2c^2+c^2a^2})^2}}{abc}$ $LHS \ge \frac{\sqrt{4abc(a+b+c)(ab+bc+ca)}}{abc}=\frac{2(ab+bc+ca)}{abc}=RHS$

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#6 h Nov 11, 2018, 7:34 AM • 2 Y

Y by Adventure10, Mango247

anhduy98 wrote:

bel.jad5 wrote:

Let $a$ , $b$ and $c$ positive real numbers such that: $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Prove that:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3 \geq 2(a+b+c)$

Use $AX-BY\ge \sqrt{A^2-B^2}.\sqrt{X^2-Y^2},(*),(A\ge B \ge 0, X\ge Y\ge 0)$ Because $(*) \Leftrightarrow (AY-BX)^2\ge 0 , (true)$ $LHS=\frac{ab^2+bc^2+ca^2+3abc}{abc}=\frac{(a+b+c)(ab+bc+ca)-(a^2b+b^2c+c^2a)}{abc}$ $LHS \ge \frac{(a+b+c)(ab+bc+ca)-\sqrt{a^2+b^2+c^2}.\sqrt{a^2b^2+b^2c^2+c^2a^2}}{abc}$ $LHS \ge \frac{\sqrt{(a+b+c)^2-(\sqrt{a^2+b^2+c^2})^2}.\sqrt{(ab+bc+ca)^2-(\sqrt{a^2b^2+b^2c^2+c^2a^2})^2}}{abc}$ $LHS \ge \frac{\sqrt{4abc(a+b+c)(ab+bc+ca)}}{abc}=\frac{2(ab+bc+ca)}{abc}=RHS$

Brilliant!!!

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sqing

#7 h Nov 26, 2020, 12:04 PM

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Let $a$ , $b$ and $c$ positive real numbers such that $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Prove that
$2(a^2+b^2+c^2)\geq a\sqrt{a^2+3}+b\sqrt{b^2+3}+c\sqrt{c^2+3}$ Let $a,b,c$ be positive real numbers, such that: $a+b+c \ge \frac {1}{a}+\frac {1}{b}+\frac{1}{c}$ . Prove that:
$a+b+c \ge \frac{3}{a+b+c}+\frac {2}{abc}$ Let $a$ , $b$ and $c$ positive real numbers such that: $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Prove that:
$a+b+c+abc \geq 4$ $\frac 1{a+2}+\frac 1{b+2}+\frac 1{c+2}\leq 1$ $\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}$ $3(a+b+c)\geq\sqrt{8a^{2}+1}+\sqrt{8b^{2}+1}+\sqrt{8c^{2}+1}.$ $(a+b+c)^{2}+3\geq 4(a+b+c)\sqrt[3]{abc}$ $(ab+bc+ca)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2 \geq 27$ https://artofproblemsolving.com/community/c6h2896381p25793992
https://artofproblemsolving.com/community/c4h2821329p25787158
https://artofproblemsolving.com/community/c4h1813778p20534203
https://artofproblemsolving.com/community/c6h355781p1932917
https://artofproblemsolving.com/community/c6h1529847p9194503
https://artofproblemsolving.com/community/c6h1739555p11304368
https://artofproblemsolving.com/community/c6h1659898p10533543

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#8 h Nov 26, 2020, 7:52 PM

Y by

nice

Stronger .
Let $a, b, c,$ be positive real numbers such that $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Prove that $a+b+c \ge abc + \frac {2}{abc} .$

This post has been edited 1 time. Last edited by Auditor, Nov 26, 2020, 7:52 PM

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#9 h Nov 26, 2020, 8:07 PM

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Auditor wrote:

nice

Stronger .
Let $a, b, c,$ be positive real numbers such that $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Prove that $a+b+c \ge abc + \frac {2}{abc} .$

Let $a, b, c,$ be non-zero real numbers such $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Prove that $a^2 + b^2 + c^2 + 6 \ge 3 (a+b+c) .$

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denery

#10 h Dec 7, 2020, 6:30 AM • 3 Y

Y by Mango247, Mango247, Mango247

@auditor im pretty sure that equality is occuring at a=b=c=1

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#11 h Dec 7, 2020, 9:13 AM • 1 Y

Y by Mango247

denery wrote:

@auditor im pretty sure that equality is occuring at a=b=c=1

And so ?

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#12 h Dec 7, 2020, 11:23 AM • 1 Y

Y by TerenceTao11235

bel.jad5 wrote:

Let $a$ , $b$ and $c$ positive real numbers such that: $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Prove that:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3 \geq 2(a+b+c)$

First,Note that $ab*(ac+bc)=ab*(1-c^2)$ so,there is no solution for $c^2>1$ (a,b,c are positive) so $0<c=<1$ and because of the symmetry $0<a=<1$ and $0<b=<1$
Thus, $0<2(a+b+c)=<6$ .
Then use AM-GM:
$a/b+b/c+c/a+3>=6$
Henceforth,
$a/b+b/c+c/a+3>=6>=2(a+b+c)$
Moreover if you use AM-HM,
$9=<(1/a+1/b+1/c)^2=(a+b+c)^2$
$3=<(a+b+c) , 6=<2(a+b+c)=<6$
So $a+b+c=3$ and because of the nature of AM-GM-HM inequalities, $(a,b,c)$ is $(1,1,1)$ .

This post has been edited 1 time. Last edited by LeonhardEuler0, Dec 7, 2020, 11:27 AM
Reason: Typo

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