Day #2 inequality 3

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Imgoodinmaths_dude

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Imgoodinmaths_dude

#1 h Jan 13, 2016, 9:17 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c>0$ and $(a+b)(b+c)(a+c)=1$ . The, prove that $ab+ac+bc\geq\frac{3}{4}$

This post has been edited 1 time. Last edited by Imgoodinmaths_dude, Jan 13, 2016, 9:17 AM

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YoungMathematicus16

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YoungMathematicus16

#2 h Jan 13, 2016, 9:39 AM • 1 Y

Y by Adventure10

ok proved, wait for a minute , i will give you solution

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Imgoodinmaths_dude

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Imgoodinmaths_dude

#3 h Jan 13, 2016, 9:41 AM • 1 Y

Y by Adventure10

YoungMathematicus16 wrote:

ok proved, wait for a minute , i will give you solution

Im waiting!

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YoungMathematicus16

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YoungMathematicus16

#4 h Jan 13, 2016, 9:57 AM • 2 Y

Y by Re1gnover, Adventure10

$((a+b)(b+c)(a+c))^{3}=1$ $(ab+bc+ac+a^{2})(b+c)(ab+bc+ac+b^{2})(a+c)(ab+bc+ac+c^{2})(a+b)=1$ let $ab+bc+ac=S$
$(S+a^{2})(S+b^{2})(S+c^{2})=1$ and $(S+a^{2})(S+b^{2})(S+c^{2})$ $\leq$ $\frac{(3S+a^{2}+b^{2}+c^{2})^{3}}{3^{3}}$ so ,
$1$ $\leq$ $\frac{3S+a^{2}+b^{2}+c^{2}}{3}$
but $S$ $\leq$ $a^{2}+b^{2}+c^{2}$ so $1$ $\leq$ $\frac{4S}{3}$ and $S$ $\geq$ $\frac{3}{4}$ and then $ab+bc+ac$ $\geq$ $\frac{3}{4}$

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Imgoodinmaths_dude

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Imgoodinmaths_dude

#6 h Jan 13, 2016, 10:04 AM • 1 Y

Y by Adventure10

YoungMathematicus16 wrote:

$((a+b)(b+c)(a+c))^{3}=1$ $(ab+bc+ac+a^{2})(b+c)(ab+bc+ac+b^{2})(a+c)(ab+bc+ac+c^{2})(a+b)=1$ let $ab+bc+ac=S$
$(S+a^{2})(S+b^{2})(S+c^{2})=1$ and $(S+a^{2})(S+b^{2})(S+c^{2})$ $\leq$ $\frac{(3S+a^{2}+b^{2}+c^{2})^{3}}{3^{3}}$ so ,
$1$ $\leq$ $\frac{3S+a^{2}+b^{2}+c^{2}}{3}$
but $S$ $\leq$ $a^{2}+b^{2}+c^{2}$ so $1$ $\leq$ $\frac{4S}{3}$ and $S$ $\geq$ $\frac{3}{4}$ and then $ab+bc+ac$ $\geq$ $\frac{3}{4}$

Thanks, what about first and second inequalities i wrote today?

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YoungMathematicus16

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YoungMathematicus16

#7 h Jan 13, 2016, 10:06 AM • 1 Y

Y by Adventure10

Imgoodinmaths_dude wrote:

YoungMathematicus16 wrote:

$((a+b)(b+c)(a+c))^{3}=1$ $(ab+bc+ac+a^{2})(b+c)(ab+bc+ac+b^{2})(a+c)(ab+bc+ac+c^{2})(a+b)=1$ let $ab+bc+ac=S$
$(S+a^{2})(S+b^{2})(S+c^{2})=1$ and $(S+a^{2})(S+b^{2})(S+c^{2})$ $\leq$ $\frac{(3S+a^{2}+b^{2}+c^{2})^{3}}{3^{3}}$ so ,
$1$ $\leq$ $\frac{3S+a^{2}+b^{2}+c^{2}}{3}$
but $S$ $\leq$ $a^{2}+b^{2}+c^{2}$ so $1$ $\leq$ $\frac{4S}{3}$ and $S$ $\geq$ $\frac{3}{4}$ and then $ab+bc+ac$ $\geq$ $\frac{3}{4}$

Thanks, what about first and second inequalities i wrote today?

ok , i will think about them

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bel.jad5

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bel.jad5

#8 h Jan 13, 2016, 10:19 AM • 2 Y

Y by Adventure10, Mango247

YoungMathematicus16 wrote:

but $S$ $\leq$ $a^{2}+b^{2}+c^{2}$ so $1$ $\leq$ $\frac{4S}{3}$ and $S$ $\geq$ $\frac{3}{4}$ and then $ab+bc+ac$ $\geq$ $\frac{3}{4}$

Are you sure about this?

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Beginer_vn86

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Beginer_vn86

#9 h Jan 13, 2016, 10:24 AM • 2 Y

Y by Adventure10, Mango247

Imgoodinmaths_dude wrote:

Let $a,b,c>0$ and $(a+b)(b+c)(a+c)=1$ . The, prove that $ab+ac+bc\geq\frac{3}{4}$

I think that $ab+bc+ca\le\frac{3}{4}.$
Indeed, it's easy to show that
$(a+b)(b+c)(c+a)\ge\frac{8}{9}(a+b+c)(ab+bc+ca)\qquad(1)$ $a+b+c\ge\sqrt{3(ab+bc+ca)}\qquad(2)$ Multiply (1) by (2) and using the given condition, we get the desired inequality.

This post has been edited 1 time. Last edited by Beginer_vn86, Jan 13, 2016, 10:25 AM

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