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

[anemone]

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

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

 

[anemone]

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

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

Hint:

Spoiler

If \(\displaystyle a\ge b \ge c\), then we would get \(\displaystyle \frac{1}{3^c}\ge \frac{1}{3^b} \ge \frac{1}{3^a}ᵃ\).

 

[anemone]

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

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

My solution:

Spoiler

WLOG, let \(\displaystyle a\ge b \ge c\) and hence \(\displaystyle \frac{1}{3^c}\ge \frac{1}{3^b} \ge \frac{1}{3^a}ᵃ\).

Apply the Chebyshev's inequality we have:

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

and therefore

\(\displaystyle \frac{1}{3^{a+1}}+\frac{1}{3^{b+1}}+\frac{1}{3^{c+1}}\ge \left(\frac{a}{3^a}+\frac{b}{3^b}+\frac{c}{3^c}\right)\) (Q.E.D.)

 

[lfdahl]

A small comment and question:

Spoiler

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 … ;)

 

[anemone]

My solution:

Spoiler

WLOG, let \(\displaystyle a\ge b \ge c\) and hence \(\displaystyle \frac{1}{3^c}\ge \frac{1}{3^b} \ge \frac{1}{3^a}ᵃ\).

Apply the Chebyshev's inequality we have:

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

and therefore

\(\displaystyle \frac{1}{3^{a+1}}+\frac{1}{3^{b+1}}+\frac{1}{3^{c+1}}\ge \left(\frac{a}{3^a}+\frac{b}{3^b}+\frac{c}{3^c}\right)\) (Q.E.D.)

Hi lfdahl(Smile),

Spoiler

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:

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

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:

\(\displaystyle n\left(\sum_{i=1}^{n}x_i y_i\right) \le \left(\sum_{i=1}^{n} x_i\right)\left(\sum_{i=1}^{n} y_i\right)\)

 

[lfdahl]

Hi, anemone!

Spoiler

Thankyou very much for the explanation. Oh, oh .. I guess, I still have a lot to learn