- #1

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if a,b,c are real numbers such that [tex]\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 2[/tex]

we have to prove that:

[tex]\frac{1}{4a+1} + \frac{1}{4b+1} + \frac{1}{4c+1} \geq 1[/tex]

thanks in advance.

- Thread starter murshid_islam
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- #1

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if a,b,c are real numbers such that [tex]\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 2[/tex]

we have to prove that:

[tex]\frac{1}{4a+1} + \frac{1}{4b+1} + \frac{1}{4c+1} \geq 1[/tex]

thanks in advance.

- #2

Gib Z

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I tried another case, where 2 solutions (a and b) are equal, and it shows a=1/(2c+1). That might help, perhaps we could prove every case to be true.

- #3

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[tex]\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1-\frac{a}{a+1}+1-\frac{b}{b+1}+1-\frac{c}{c+1}=2[/tex]

Hence,the condition is equivalent with:

[tex]\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=1[/tex]

By similar simple algebra the inequality can be rewritten as follows:

[tex]\frac{4a}{4a+1}+\frac{4b}{4b+1}+\frac{4c}{4c+1}\leq 2 [/tex]

Now,let's consider the difference "

[tex]\left(\frac{4a}{4a+1}-\frac{a}{a+1}\right)+\left(\frac{4b}{4b+1}-\frac{b}{b+1}\right)+\left(\frac{4c}{4c+1}-\frac{b}{b+1}\right)\leq 1[/tex]

.

Every expression within brackets is of form [itex]\frac{4x}{4x+1}-\frac{x}{x+1}[/itex].

[tex]\frac{4x}{4x+1}-\frac{x}{x+1}>\frac{1}{3}?\rightarrow 4x^2-4x+1<0\rightarrow 4\left(x-\frac{1}{2}\right)^2<0\longrightarrow Contradiction![/tex]

EDIT:BTW,this cute little problem was misplaced in dumb brainteasers section.

Elementar proof presented here is understandable even to 6

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- #4

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That's simply a brilliant solution. Wow !

- #5

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how do you get a contradiction here? i get:[tex]\frac{4x}{4x+1}-\frac{x}{x+1}>\frac{1}{3}?\rightarrow 4x^2-4x+1<0\rightarrow 4\left(x-\frac{1}{2}\right)^2<0\longrightarrow Contradiction![/tex]

[tex]\frac{4x}{4x+1} - \frac{x}{x+1} > \frac{1}{3}[/tex]

[tex]\frac{3x}{(4x+1)(x+1)} > \frac{1}{3}[/tex]

[tex]\frac{9x}{(4x+1)(x+1)} > 1[/tex]

[tex]\frac{9x}{(4x+1)(x+1)} - 1 > 0[/tex]

[tex]\frac{9x - 4x^2 - 5x - 1}{(4x+1)(x+1)} > 0[/tex]

[tex]\frac{4x^2 - 4x + 1}{(4x+1)(x+1)} < 0[/tex]

[tex]\frac{(2x-1)^2}{(4x+1)(x+1)} < 0[/tex]

[tex](4x+1)(x+1) < 0[/tex]

[tex]x \in \left(-\frac{1}{4}, -1\right)[/tex]

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- #6

Gib Z

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Take a look at the original question, a b and c have to be POSTIVE. All the solutions to that quadratic inequality are negative, and therefore we reach a contradiction.

- #7

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thanks Gib Z. but still, the contradiction tehno got is wrong, isn't it?

- #8

Gib Z

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- #9

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how do you get from [tex]\frac{4x}{4x+1}-\frac{x}{x+1}>\frac{1}{3}[/tex] to [tex]4x^2-4x+1<0[/tex]tehno said:[tex]\frac{4x}{4x+1}-\frac{x}{x+1}>\frac{1}{3}?\rightarrow 4x^2-4x+1<0[/tex]

thats what i said was wrong?

- #10

Gib Z

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Well ive tried for 30 mins now, i cant see how he got that either. Help tehno!

- #11

cristo

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From here, x is positive, and so the denominator is always positive. So, for the inequality to hold, we require [itex]4x^2 - 4x + 1<0[/itex]. From here, the contradiction is obtained as in tehno's post.how do you get a contradiction here? i get:

[tex]\frac{4x}{4x+1} - \frac{x}{x+1} > \frac{1}{3}[/tex]

[tex]\frac{3x}{(4x+1)(x+1)} > \frac{1}{3}[/tex]

[tex]\frac{9x}{(4x+1)(x+1)} > 1[/tex]

[tex]\frac{9x}{(4x+1)(x+1)} - 1 > 0[/tex]

[tex]\frac{9x - 4x^2 - 5x - 1}{(4x+1)(x+1)} > 0[/tex]

[tex]\frac{4x^2 - 4x + 1}{(4x+1)(x+1)} < 0[/tex]

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