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Positive real number inequality

conscipost

Member
Jan 26, 2012
39
Prove that for positive real numbers a,b (a+1/b+1)^(b+1) is greater than or equal to (a/b)^(b).
The case in which a<b is easy to prove, but after trying to represent the inequality with an integral, I'm a bit stumped.

Any ideas?
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Let $b$ be a fixed positive number. Define the following function for $x>0$:

$$ f(x) = \frac{(x+1)^{b+1}}{x^b} $$

Let us find the critical points of $f$ (the points where the derivative is zero). By the quotient rule the numerator of the derivative is equal to (we will ignore the denominator, as you will see why):

$$ \left[ (x+1)^{b+1} \right]' x^b - (x+1)^{b+1} \left[ x^b \right]' $$

This becomes,

$$ (b+1)(x+1)^b x^b - b(x+1)^{b+1}x^{b-1} $$

This is the numerator for $f'(x)$. To find the critical points of $f$ we just need to set the numerator equal to zero:

$$ (b+1)(x+1)^b x^b - b(x+1)^{b+1}x^{b-1} = 0$$

Divide both sides by $(x+1)^bx^{b-1}$ to end up with:

$$ (b+1)x - b(x+1) = 0 \implies bx + x - bx - b = 0 \implies x = b$$

Therefore, as $b$ is the critical point it means $f$ is either increasing or decreasing on the interval $(0,b)$ and that it is either increasing or decreasing on the interval $(b,\infty)$. Note that,

$$ \lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} \frac{(x+1)^{b+1}}{x^b} = \infty $$

This means that $f$ must be decreasing on $(0,b)$.

Also note that,

$$ \lim_{x\to \infty} f(x) = \lim_{x\to \infty} \frac{(x+1)^{b+1}}{x^b} = \infty $$

Therefore, $f$ must be increasing on $(b,\infty)$.

This means that the point $b$ is a global minimum point for $f$. Thus, we have that $f(x) \geq f(b)$ for any point $x$. Now choose $x=a$ where $a$ is any positive number and we get that,

$$ f(a) \geq f(b) \implies \frac{(a+1)^{b+1}}{a^b} \geq \frac{(b+1)^{b+1}}{b^b} \implies \frac{(a+1)^{b+1}}{(b+1)^{b+1}} \geq \frac{a^b}{b^b} \implies \left( \frac{a+1}{b+1} \right)^{b+1} \geq \left( \frac{a}{b} \right)^b $$
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Here's another way. You've already said that

$\left(\dfrac{a}{b}\right)^b \le \left( \dfrac{a+1}{b+1}\right)^{b+1}$ for $a < b$ is easy to prove - right?

if $a = b$, this is also easy since $1 \le 1$. Here's the case where $b < a$. Set $a \to \dfrac{1}{a}$ and $b \to \dfrac{1}{b}$

so $\left( \dfrac{\dfrac{1}{a}}{\dfrac{1}{b}}\right)^{\dfrac{1}{b}} \le \left(\dfrac{\dfrac{1}{a}+1}{\dfrac{1}{b}+1} \right) ^{\dfrac{1}{b}+1}$ if $\dfrac{1}{a} < \dfrac{1}{b}$

Now we simplify

$ \left(\dfrac{b}{a}\right)^{\dfrac{1}{b}} \le \left(\dfrac{b}{a}\right)^{\dfrac{1}{b}+1} \left(\dfrac{a+1}{b+1}\right)^{\dfrac{1}{b}+1}$

cancel some

$ 1 \le \dfrac{b}{a} \left(\dfrac{a+1}{b+1}\right)^{\dfrac{b+1}{b}}$ or $ \dfrac{a}{b} \le \left(\dfrac{a+1}{b+1}\right)^{\dfrac{b+1}{b}}$

Now exponenate each side giving

$ \left(\dfrac{a}{b}\right)^b \le \left(\dfrac{a+1}{b+1}\right)^{b+1}$ for $b < a$