# A new problem

murshid_islam
the problem statement is:
if a,b,c are real numbers such that $$\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 2$$

we have to prove that:
$$\frac{1}{4a+1} + \frac{1}{4b+1} + \frac{1}{4c+1} \geq 1$$

Homework Helper
Well I thought id at least prove a general class of the solutions, where a=b=c, but it turns out that the only solution is a=b=c=0.5. It is the only solution where the 2nd equtation is exactly 1. The others must be more.

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.

tehno
Condition:
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1-\frac{a}{a+1}+1-\frac{b}{b+1}+1-\frac{c}{c+1}=2$$

Hence,the condition is equivalent with:
$$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=1$$

Inequality:
By similar simple algebra the inequality can be rewritten as follows:
$$\frac{4a}{4a+1}+\frac{4b}{4b+1}+\frac{4c}{4c+1}\leq 2$$

Now,let's consider the difference "Inequality-Condition":

$$\left(\frac{4a}{4a+1}-\frac{a}{a+1}\right)+\left(\frac{4b}{4b+1}-\frac{b}{b+1}\right)+\left(\frac{4c}{4c+1}-\frac{b}{b+1}\right)\leq 1$$
.

Every expression within brackets is of form $\frac{4x}{4x+1}-\frac{x}{x+1}$.

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

QED

EDIT:BTW,this cute little problem was misplaced in dumb brainteasers section.
Elementar proof presented here is understandable even to 6th graders.

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Homework Helper
That's simply a brilliant solution. Wow !

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

$$\frac{4x}{4x+1} - \frac{x}{x+1} > \frac{1}{3}$$

$$\frac{3x}{(4x+1)(x+1)} > \frac{1}{3}$$

$$\frac{9x}{(4x+1)(x+1)} > 1$$

$$\frac{9x}{(4x+1)(x+1)} - 1 > 0$$

$$\frac{9x - 4x^2 - 5x - 1}{(4x+1)(x+1)} > 0$$

$$\frac{4x^2 - 4x + 1}{(4x+1)(x+1)} < 0$$

$$\frac{(2x-1)^2}{(4x+1)(x+1)} < 0$$

$$(4x+1)(x+1) < 0$$

$$x \in \left(-\frac{1}{4}, -1\right)$$

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Homework Helper
Ahh here's your problem murshid_islam, you know when you copyed this problem over from the brain teasers section? https://www.physicsforums.com/showthread.php?t=131366

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.

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

Homework Helper
Why is the contradiction wrong? The inequaility has only negative solutions, which is contradictory from the original assumption of positive solutions.

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

thats what i said was wrong?

Homework Helper
Well I've tried for 30 mins now, i can't see how he got that either. Help tehno!

Staff Emeritus
how do you get a contradiction here? i get:

$$\frac{4x}{4x+1} - \frac{x}{x+1} > \frac{1}{3}$$

$$\frac{3x}{(4x+1)(x+1)} > \frac{1}{3}$$

$$\frac{9x}{(4x+1)(x+1)} > 1$$

$$\frac{9x}{(4x+1)(x+1)} - 1 > 0$$

$$\frac{9x - 4x^2 - 5x - 1}{(4x+1)(x+1)} > 0$$

$$\frac{4x^2 - 4x + 1}{(4x+1)(x+1)} < 0$$

From here, x is positive, and so the denominator is always positive. So, for the inequality to hold, we require $4x^2 - 4x + 1<0$. From here, the contradiction is obtained as in tehno's post.

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