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

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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.
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Let $a,\,b$ and $c$ be real numbers such that $a+b+c=1$, prove that

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