MHB Prove Inequality Problem for Real Numbers $a, b, c$ with $a + b + c = 1$

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Inequality
AI Thread Summary
The discussion focuses on proving the inequality involving real numbers \(a\), \(b\), and \(c\) such that \(a + b + c = 1\). The inequality states that \(\frac{1}{3^{a+1}} + \frac{1}{3^{b+1}} + \frac{1}{3^{c+1}} \ge \left(\frac{a}{3^a} + \frac{b}{3^b} + \frac{c}{3^c}\right)\). Participants are encouraged to explore potential methods for proving this inequality, including hints and approaches. The conversation includes a mix of solutions and clarifications regarding the mathematical expressions involved. The thread emphasizes the importance of understanding the relationship between the variables in the context of the given constraint.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Let $a,\,b$ and $c$ be real numbers such that $a+b+c=1$, prove that

$$\frac{1}{3^{a+1}}+\frac{1}{3^{b+1}}+\frac{1}{3^{c+1}}\ge \left(\frac{a}{3^a}+\frac{b}{3^b}+\frac{c}{3^c}\right)$$.
 
Mathematics news on Phys.org
anemone said:
Let $a,\,b$ and $c$ be real numbers such that $a+b+c=1$, prove that

$$\frac{1}{3^{a+1}}+\frac{1}{3^{b+1}}+\frac{1}{3^{c+1}}\ge \left(\frac{a}{3^a}+\frac{b}{3^b}+\frac{c}{3^c}\right)$$.

Hint:

If $$a\ge b \ge c$$, then we would get $$\frac{1}{3^c}\ge \frac{1}{3^b} \ge \frac{1}{3^a}ᵃ$$.
 
anemone said:
Let $a,\,b$ and $c$ be real numbers such that $a+b+c=1$, prove that

$$\frac{1}{3^{a+1}}+\frac{1}{3^{b+1}}+\frac{1}{3^{c+1}}\ge \left(\frac{a}{3^a}+\frac{b}{3^b}+\frac{c}{3^c}\right)$$.

My solution:
WLOG, let $$a\ge b \ge c$$ and hence $$\frac{1}{3^c}\ge \frac{1}{3^b} \ge \frac{1}{3^a}ᵃ$$.

Apply the Chebyshev's inequality we have:

$$\begin{align*}3\left(\frac{a}{3^a}+\frac{b}{3^b}+\frac{c}{3^c}\right)&\le \left(a+b+c\right)\left(\frac{1}{3^a}+\frac{1}{3^b}+\frac{1}{3^c}\right)\\&\le \left(1\right)\left(\frac{1}{3^a}+\frac{1}{3^b}+\frac{1}{3^c}\right)\end{align*}$$

and therefore

$$\frac{1}{3^{a+1}}+\frac{1}{3^{b+1}}+\frac{1}{3^{c+1}}\ge \left(\frac{a}{3^a}+\frac{b}{3^b}+\frac{c}{3^c}\right)$$ (Q.E.D.)
 
A small comment and question:

Hi, anemone

Thankyou so much for your indefatigable contributions to this interesting site!(Yes)
You, MarkFL et al. are really doing a great job. Thankyou so much!I have a small comment/question to this challenge.Maybe the inequality is not Chebyshev (statistics), but Cauchy-Schwarz (inner product)??Even if so, I do not understand how to use it in the context. If I define two vectors:\[\bar{u} = \begin{pmatrix} a\\ b\\ c \end{pmatrix}, \: \: \: \bar{v} = \begin{pmatrix} \ 1/3^a\\ 1/3^b \\ 1/3^c \end{pmatrix}\]then using the Cauchy-Schwarz inequality:

\[\left ( \sum_{i=1}^{n} u_iv_i\right )^2 \leq \left ( \sum_{j=1}^{n}u_j^2 \right )\left ( \sum_{k=1}^{n} v_k^2\right )\]

- I get:

\[\left ( \frac{a}{3^a}+\frac{b}{3^b}+\frac{c}{3^c} \right )^2 \leq \left ( a^2+b^2+c^2 \right )\left ( \frac{1}{3^{2a}} + \frac{1}{3^{2b}}+ \frac{1}{3^{2c}}\right )\]How can I ommit squaring?Thankyou in advance for clearing my confusion … ;)
 
Last edited:
anemone said:
My solution:
WLOG, let $$a\ge b \ge c$$ and hence $$\frac{1}{3^c}\ge \frac{1}{3^b} \ge \frac{1}{3^a}ᵃ$$.

Apply the Chebyshev's inequality we have:

$$\begin{align*}3\left(\frac{a}{3^a}+\frac{b}{3^b}+\frac{c}{3^c}\right)&\le \left(a+b+c\right)\left(\frac{1}{3^a}+\frac{1}{3^b}+\frac{1}{3^c}\right)\\&\le \left(1\right)\left(\frac{1}{3^a}+\frac{1}{3^b}+\frac{1}{3^c}\right)\end{align*}$$

and therefore

$$\frac{1}{3^{a+1}}+\frac{1}{3^{b+1}}+\frac{1}{3^{c+1}}\ge \left(\frac{a}{3^a}+\frac{b}{3^b}+\frac{c}{3^c}\right)$$ (Q.E.D.)
Hi lfdahl(Smile),

The inequality formula that I used is the Chebyshev's inequality that says:

If $x_1\ge x_2\ge \cdots\ge x_n$ and $y_1\ge y_2\ge \cdots\ge y_n$, then the following inequality holds:

$$n\left(\sum_{i=1}^{n}x_i y_i\right) \ge \left(\sum_{i=1}^{n} x_i\right)\left(\sum_{i=1}^{n} y_i\right)$$

On the other hand, if $x_1\ge x_2\ge \cdots\ge x_n$ and $y_n\ge y_{n-1}\ge \cdots\ge y_1$ then we have:

$$n\left(\sum_{i=1}^{n}x_i y_i\right) \le \left(\sum_{i=1}^{n} x_i\right)\left(\sum_{i=1}^{n} y_i\right)$$
 
Hi, anemone!

Thankyou very much for the explanation. Oh, oh .. I guess, I still have a lot to learn :o
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Replies
1
Views
861
Replies
10
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
3
Views
1K
Replies
2
Views
2K
Back
Top