Proving Real Numbers Satisfy Inequality

[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:
$$\frac{1}{4a+1} + \frac{1}{4b+1} + \frac{1}{4c+1} \geq 1$$

thanks in advance.

 

[Gib Z]

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 6^th graders.

 

[dextercioby]

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:

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

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

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

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

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

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

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

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

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

so where is the contradiction?

 

[Gib Z]

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 positive. 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?

 

[Gib Z]

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

 

[murshid_islam]

$$\frac{4x}{4x+1}-\frac{x}{x+1}>\frac{1}{3}?\rightarrow 4x^2-4x+1<0$$

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?

 

[Gib Z]

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

 

[cristo]

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.