Let $a, b$ and $c$ be non-zero natural numbers such that $c \geq b$ . Show that $$a^b\left(a+b\right)^c>c^b a^c.$$
Problem
Source: St. Petersburg Mathematical Olympiad 2019
Tags: inequalities
ThE-dArK-lOrD
13.04.2019 11:30
Try $(a,b,c)=(1,6,2)$.
XbenX
13.04.2019 11:53
In the original problem there is the condition $c \geq b$ and this is the 10th grade Problem 3.
sqing
13.04.2019 12:57
ThE-dArK-lOrD wrote: Try $(a,b,c)=(1,6,2)$. Sorry.
XbenX
13.04.2019 14:23
sqing wrote: Let $a, b$ and $c$ be non-zero natural numbers such that $c \geq b$ . Show that $$a^b\left(a+b\right)^c>c^b a^c.$$ Note that the both sides being positive integers we can take the $b$'th root and still have the same sign so doing this we have $$a(a+b)^{\frac{c}{b}}>c \cdot a^{\frac{c}{b}} \Longleftrightarrow \left ( 1+\frac{b}{a}\right)^{\frac{c}{b}}>\frac{c}{a}$$ Since $\frac{c}{b} \geq 1$ we apply Bernoulli inequality to get $$\left ( 1+\frac{b}{a}\right)^{\frac{c}{b}}\geq 1+\frac{b}{a} \cdot \frac{c}{b}=1+\frac{c}{a}>\frac{c}{a}$$$\blacksquare$.
sqing
13.04.2019 14:25
XbenX wrote:
sqing wrote:
Let $a, b$ and $c$ be non-zero natural numbers such that $c \geq b$ . Show that
$$a^b\left(a+b\right)^c>c^b a^c.$$
Note that the both sides being positive integers we can take the $b$'th root and still have the same sign so doing this we have $$a(a+b)^{\frac{c}{b}}>c \cdot a^{\frac{c}{b}} \Longleftrightarrow \left ( 1+\frac{b}{a}\right)^{\frac{c}{b}}>\frac{c}{a}$$
Since $\frac{c}{b} \geq 1$ we apply Bernoulli inequality to get $$\left ( 1+\frac{b}{a}\right)^{\frac{c}{b}}\geq 1+\frac{b}{a} \cdot \frac{c}{b}=1+\frac{c}{a}>\frac{c}{a}$$$\blacksquare$.