If a,b,c in (0,+infty), abc=1, prove a/(2+bc)+b/(2+ac)+c/(2+ab)>=1

[Roger.Robert]

If \(\displaystyle \displaystyle a,\, b\, c\, \in\, (0,\, +\infty)\) and a*b*c=1, prove that:

\(\displaystyle \displaystyle \frac{a}{2\,+\,b*c}\,+\,\frac{b}{2\,+\,a*c}\,+\, \frac{c}{2\,+\,a*b} \,\ge\, 1\)

so

\(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\,\geq\, 1\)

I don't know what to do from here..​

 

[stapel]

If \(\displaystyle \displaystyle a,\, b\, c\, \in\, (0,\, +\infty)\) and a*b*c=1, prove that:

\(\displaystyle \displaystyle \frac{a}{2\,+\,b*c}\,+\,\frac{b}{2\,+\,a*c}\,+\, \frac{c}{2\,+\,a*b} \,\ge\, 1\)

so

\(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\,\geq\, 1\)

I don't know what to do from here..​

Are you saying that you got the second line of fractions from the first line of fractions, and don't know what to do next?

 

[Steven G]

If \(\displaystyle \displaystyle a,\, b\, c\, \in\, (0,\, +\infty)\) and a*b*c=1, prove that:

\(\displaystyle \displaystyle \frac{a}{2\,+\,b*c}\,+\,\frac{b}{2\,+\,a*c}\,+\, \frac{c}{2\,+\,a*b} \,\ge\, 1\)

so

\(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\,\geq\, 1\)

I don't know what to do from here..​

maybe get common denominators.

 

[Roger.Robert]

Are you saying that you got the second line of fractions from the first line of fractions, and don't know what to do next?

yes

 

[Steven G]

yes

Maybe try my hint from the other post??

 

[Steven G]

If \(\displaystyle \displaystyle a,\, b\, c\, \in\, (0,\, +\infty)\) and a*b*c=1, prove that:

\(\displaystyle \displaystyle \frac{a}{2\,+\,b*c}\,+\,\frac{b}{2\,+\,a*c}\,+\, \frac{c}{2\,+\,a*b} \,\ge\, 1\)

so

\(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\,\geq\, 1\)

I don't know what to do from here..​

Actually I find it hard to believe that you were able to get to where you are but do not know how to continue. Can you explain what you did to get to this 2nd line. Thank you.

 

[JeffM]

Actually I find it hard to believe that you were able to get to where you are but do not know how to continue. Can you explain what you did to get to this 2nd line. Thank you.

Jomo

\(\displaystyle abc = 1 \implies bc = \dfrac{1}{a} \implies \dfrac{a}{2 + bc} = \dfrac{a}{2 + \dfrac{1}{a}} * \dfrac{a}{a} = \dfrac{a^2}{2a + 1}.\)

And so on for the rest of the fractions.

I have played around with this problem from the second line, and a straight algebraic solution seems quite messy on that basis. One idea that I did not explore is to set a = 1 + d, b = 1 + e, c = 1 + f. You can assume without loss of generality that

\(\displaystyle 0 < a \le b \le c \implies -\ 1 < d \le 0 \text { and } 0 \le f.\)

LATE EDIT: For some reason, some less than or equal signs are showing as less than.

LATER EDIT: Now the signs are rendering properly

 

[Steven G]

Jomo

\(\displaystyle abc = 1 \implies bc = \dfrac{1}{a} \implies \dfrac{a}{2 + bc} = \dfrac{a}{2 + \dfrac{1}{a}} * \dfrac{a}{a} = \dfrac{a^2}{2a + 1}.\)

And so on for the rest of the fractions.

I have played around with this problem from the second line, and a straight algebraic solution seems quite messy on that basis. One idea that I did not explore is to set a = 1 + d, b = 1 + e, c = 1 + f. You can assume without loss of generality that

\(\displaystyle 0 < a \le b \le c \implies -\ 1 < d \le 0 \text { and } 0 \le f.\)

LATE EDIT: For some reason, some less than or equal signs are showing as less than.

LATER EDIT: Now the signs are rendering properly

JeffM, thanks for the reply. I see how the OP got to the 2nd line, I just have a problem believing that he got there on his own given that he does not know how to continue at all.

 

[Roger.Robert]

I managed to solve the problem,here's how i did it.
Prove that the function \(\displaystyle \displaystyle \frac{x^{2}}{2*x\,+\,1}\) is increasing on the \(\displaystyle \displaystyle (0,\, +\infty)\) interval by comparing increasing values of x to the value of the function.
All three functions \(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,,\,\frac{b^{2}}{2*b\,+\,1}\,,\,\frac{c^{2}}{2*c\,+\,1}\) increase at the same rate (because they are the same function)\(\displaystyle \Large\Rightarrow\) the value of the function \(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\) is equally dependant on the three variables \(\displaystyle \Large\Rightarrow\).
\(\displaystyle abc=1\) \(\displaystyle \Large\Rightarrow\) at least one of the variables must be equal or bigger than 1.
\(\displaystyle abc=x^{-1}x=1\) ,smallest value for x>= 1 is 1;\(\displaystyle \Large\Rightarrow\) a=1 b=1 c=1
We can conclude that the minimum value of \(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\) is 1 with a=b=c=1;

 

[JeffM]

I have not worked this out, in part because I am busy with other things and in part because we have no clue what the OP is studying: this is certainly not beginning algebra where it was posted.

\(\displaystyle 0 < a \le b \le c \implies 0 < \sqrt[3]{abc} \le \dfrac{a}{3} + \dfrac{b}{3} + \dfrac{c}{3}.\)

Does the OP know this? How do we know when the OP gives us no clue about the context of the problem?

\(\displaystyle 0 < a \le b \le c \text { and } abc = 1 \implies abc \le \dfrac{1}{27} * ( a + b + c)^3.\)

As I said, I have not had time to work on this so I don't know whether it is a helpful clue.

 

[JeffM]

Here's how i solved it.
Prove that the function \(\displaystyle \displaystyle \frac{x^{2}}{2*x\,+\,1}\) is increasing on the \(\displaystyle \displaystyle (0,\, +\infty)\) interval by comparing increasing values of x to the value of the function.
All three functions \(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,,\,\frac{b^{2}}{2*b\,+\,1}\,,\,\frac{c^{2}}{2*c\,+\,1}\) increase at the same rate (because they are the same function)\(\displaystyle \Large\Rightarrow\) the value of the function \(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\) is equally dependant on the three variables \(\displaystyle \Large\Rightarrow\).
\(\displaystyle abc=1\) \(\displaystyle \Large\Rightarrow\) at least one of the variables must be equal or bigger than 1.
\(\displaystyle abc=x^{-1}x=1\) ,smallest value for x>= 1 is 1;\(\displaystyle \Large\Rightarrow\) a=1 b=1 c=1
We can conclude that the minimum value of \(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\) is 1 with a=b=c=1;

Sorry for the bad english.

Edit:I made a mistake.

You certainly did make a mistake.

\(\displaystyle \dfrac{x^2}{2x + 1}\) is a function becaus x is a variable.

Its rate of increase is \(\displaystyle \dfrac{4x^2 +2x - 2x^2}{(2x + 1)^2} = \dfrac{2x(x + 1)}{(2x + 1))^2}.\)

But \(\displaystyle \dfrac{a^2}{2a + 1}\) is a number, not a function, because a is a number.

in any case, you cannot increase a without decreasing b or c or both because their product is fixed. Consequently, the fact that the rate of change is positive seems irrelevant.

Jomo and I have both given you suggestions. Play around with them.

 

[Roger.Robert]

Actually there is a solution at the end of the textbook from where the problem is:
"The left member of the inequality is written\(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\,\geq\ \frac{(a+b+c)^{2}}{2(a+b+c)+3}\).It also needs to be shown that\(\displaystyle \displaystyle \frac{(a+b+c)^{2}}{2(a+b+c)+3}\, \ge 1\) \(\displaystyle \Large\Leftrightarrow\) \(\displaystyle (a+b+c-1)^{2}\, \ge 4\).But \(\displaystyle \displaystyle a+b+c\, \ge 3\) and \(\displaystyle \displaystyle a+b+c-1\, \ge 2\) , therefore \(\displaystyle \displaystyle (a+b+c-1)^{2}\, \ge 4 \)"
But i don't get where does \(\displaystyle \displaystyle \frac{(a+b+c)^{2}}{2(a+b+c)+3}\) come from and why it is smaller or equal than \(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\).



//////////////////////////////////
Btw this problem is from a 9th grade textbook(i'm 11th grade)

 

[Roger.Robert]

You certainly did make a mistake.

\(\displaystyle \dfrac{x^2}{2x + 1}\) is a function becaus x is a variable.

Its rate of increase is \(\displaystyle \dfrac{4x^2 +2x - 2x^2}{(2x + 1)^2} = \dfrac{2x(x + 1)}{(2x + 1))^2}.\)

But \(\displaystyle \dfrac{a^2}{2a + 1}\) is a number, not a function, because a is a number.

in any case, you cannot increase a without decreasing b or c or both because their product is fixed. Consequently, the fact that the rate of change is positive seems irrelevant.

Jomo and I have both given you suggestions. Play around with them.

Couldn't i say that f=\(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\) is a multivariable function with three variables: (a,b,c)?
I want to prove that the function has no smaller value than 1 for the condition \(\displaystyle abc=1\) (In other words the function minimum is 1 for that condition or the smallest value of the three variables).
It is useful to know that the functions \(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,,\,\frac{b^{2}}{2*b\,+\,1}\,,\,\frac{c^{2}}{2*c\,+\,1}\) increase at the same rate as the three variables have the same impact (probably not the right word) on the bigger function.
To meet the condition \(\displaystyle abc=1\) in \(\displaystyle \displaystyle (0,\, +\infty)\)
1)\(\displaystyle b=a^{-1}\) and \(\displaystyle c=1 \)
2) \(\displaystyle c=(ab)^{-1}\)
The smallest value of a,b,c that meets the condition (or the smallest value of a+b+c that meets the condition) is 3 with a=b=c=1 and \(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1} =1\)
It makes sense to me..

 

[JeffM]

Couldn't i say that f=\(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\) is a multivariable function with three variables: (a,b,c)?
I want to prove that the function has no smaller value than 1 for the condition \(\displaystyle abc=1\) (In other words the function minimum is 1 for that condition or the smallest value of the three variables).
It is useful to know that the functions \(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,,\,\frac{b^{2}}{2*b\,+\,1}\,,\,\frac{c^{2}}{2*c\,+\,1}\) increase at the same rate as the three variables have the same impact (probably not the right word) on the bigger function.

They contribute the same amount to the function if a, b, and c are all equal. That is not something that you are given.

To meet the condition \(\displaystyle abc=1\) in \(\displaystyle \displaystyle (0,\, +\infty)\)
1)\(\displaystyle b=a^{-1}\) and \(\displaystyle c=1 \)
2) \(\displaystyle c=(ab)^{-1}\)

But that is a special case. You are not given that c = 1.

The smallest value of a,b,c that meets the condition (or the smallest value of a+b+c that meets the condition) is 3 with a=b=c=1 and \(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1} =1\)
It makes sense to me..

You can easily prove it with calculus by setting up a function with three variables and using a LaGrangian multiplier. But are you supposed to do that? You posted this in beginning algebra.

\(\displaystyle z = \dfrac{a^2}{2a + 1} + \dfrac{b^2}{2b + 1} + \dfrac{c^2}{2c + 1}.\)

Minimize z subject to the constraint that \(\displaystyle abc = 1.\)

Note that now we are treating a, b, and c as variables, not numbers.

\(\displaystyle O = \dfrac{a^2}{2a + 1} + \dfrac{b^2}{2b + 1} + \dfrac{c^2}{2c + 1} + \lambda (1 - abc) \implies\)

\(\displaystyle \dfrac{\delta O}{\delta a} = 0 \iff \dfrac{2a(a + 1)}{(2a + 1)^2} - \lambda = 0 \implies \dfrac{2a(a + 1)}{(2a + 1)^2} = \lambda\)

\(\displaystyle \dfrac{\delta O}{\delta b} = 0 \iff \dfrac{2b(b + 1)}{(2b + 1)^2} - \lambda = 0 \implies \dfrac{2b(b + 1)}{(2b + 1)^2} = \dfrac{2a(a + 1)}{(2a + 1)^2} \implies\)

\(\displaystyle (2b^2 + 2b)(4a^2 + 4a + 1) = (2a^2 + 2a)(4b^2 + 4b + 1) \implies\)

\(\displaystyle 8a^2b^2 + 8ab^2 + 2b^2 + 8a^2b + 8ab + 2b = 8a^2b^2 + 8a^2b + 2a^2 +8ab^2 + 8ab + 2a \implies\)

\(\displaystyle 8a^2b^2 - 8a^2b^2 + 8ab^2 - 8ab^2 + 2b^2 - 2a^2 + 8a^2b - 8a^2b + 8ab - 8ab + 2b - 2a = \implies\)

\(\displaystyle 2b^2 - 2a^2 2b - 2a = 0 \implies b^2 - a^2 + b - a = 0 \implies (b - a)(b + a) + (b - a)(1) = 0 \implies\)

\(\displaystyle (b - a)(b + a + 1) = 0 \implies a = b \ \because a > 0 < b \text { by hypothesis.}\)

Similarly, you can prove that a = c so a = b = c so a, b, and c all equal 1.

To show then that 1 is a minimum you need to look at second partials.

EDIT: But this is not an algebraic demonstration. This is why it is important to say what you are studying, what you know, and how you are supposed to do a problem. We cannot guess those things.

 

[Steven G]

Actually there is a solution at the end of the textbook from where the problem is:
"The left member of the inequality is written\(\displaystyle \displaystyle \frac{a^{2}}{2*a\,+\,1}\,+\,\frac{b^{2}}{2*b\,+\,1}\,+\,\frac{c^{2}}{2*c\,+\,1}\,\geq\ \frac{(a+b+c)^{2}}{2(a+b+c)+3}\).