$\sum_{1}^{n}\left ( a_{i}b_{i} \right )\geq \prod_{1}^{n}\left ( a_{i}^{b_{i}} \right )$

Question

Let $a,b,c,x,y,z>0$ and $x+ y+ z= 1$. Show that $$ax+ by+ cz\geq a^{x}.b^{y}.c^{z}$$ I have found the generalization of this inequality: For 2 sets of numbers $\left \{ a_{i} \right \}, \left \{ b_{i} \right \}> 0$ and $\sum_{1}^{n}b_{i}= 1$, we have: $$\sum_{1}^{n}\left ( a_{i}b_{i} \right )\geq \prod_{1}^{n}\left ( a_{i}^{b_{i}} \right )$$ First solution Simply compare the logs: as the natural logarithm is a concave function, one has $$\ln\bigl(a^x b^y c^z)=x\ln a+y\ln b+z\ln c\le \ln(ax+by+cz)$$ for all non-negative $x,y,z$ such that $x+y+z=1$. How about another solution? I really hope to see that. Thanks!

Answer 1

Another way to prove it is starting with the AM-GM inequality: $$ \frac{1}{n}(x_1 + x_2 + \dots + x_n) \geq (x_1 x_2 \cdots x_n)^{\frac{1}{n}} $$ for all positive real numbers $x_1, x_2, \dots, x_n$.

Suppose that we have rational numbers $a_1, a_2, \dots, a_n$ such that $a_1 + a_2 + \dots + a_n = 1$, and some positive real numbers $x_1, x_2, \dots, x_n$. We will show that $$ a_1 x_1 + a_2 x_2 + \dots + a_n x_n \geq {x_1}^{a_1} {x_2}^{a_2} \cdots {x_n}^{a_n}, $$ and so your inequality holds when the weights are rational.

To see this, let $N$ be the least common multiple of the denominators of the $a_i$. Then we can write $$ a_i = \frac{m_i}{N} $$ for each $i$. The AM-GM inequality then tells us that

$$ a_1 x_1 + a_2 x_2 + \dots + a_n x_n = \frac{1}{N} (m_1 x_1 + m_2 x_2 + \dots + m_n x_n) = \frac{1}{N} (\overbrace{x_1 + x_1 + \dots + x_1}^{m_1 \text{ times}} + \overbrace{x_2 + x_2 + \dots + x_2}^{m_2 \text{ times}} + \dots + \overbrace{x_n + x_n + \dots + x_n}^{m_n \text{ times}}) \geq \left( {x_1}^{m_1} {x_2}^{m_2} \cdots {x_n}^{m_n} \right)^{\frac{1}{N}} = {x_1}^{a_1} {x_2}^{a_2} \cdots {x_n}^{a_n}. $$

We now extend the inequlity to all real weights by using a limiting argument.

Consider now any positive real numbers $a_1, a_2, \dots, a_n$ such that $a_1 + a_2 + \dots + a_n = 1$.

For each $i$ from $1$ to $n - 1$, we can find an increasing sequence of positive rational numbers $$ w_1^{(i)}, w_2^{(i)}, w_3^{(i)}, \dots $$ such that $$ \lim_{k \to \infty} w_k^{(i)} = a_i. $$

Let $$ w_k^{(n)} = 1 - w_k^{(1)} - w_k^{(2)} - \dots - w_{k}^{(n-1)} $$ for each $k$. Then for each $k$, we have that $w_k^{(n)}$ is also a positive rational number, and that $$ \lim_{k \to \infty} w_k^{(n)} = a_n. $$

Using our result from earlier, we know that for each $k$, we have that $$ w_k^{(1)} x_1 + w_k^{(2)} x_2 + \dots + w_k^{(n)} x_n \geq {x_1}^{w_k^{(1)}} {x_2}^{w_k^{(2)}} \cdots {x_n}^{w_k^{(n)}}. $$

Taking the limit as $k \to \infty$ then gives us that $$ a_1 x_1 + a_2 x_2 + \dots + a_n x_n \geq {x_1}^{a_1} {x_2}^{a_2} \cdots {x_n}^{a_n}. $$