Question 1209906: Let a and b be positive real numbers. Find the minimum value of
(a + b) \left( \frac{1}{a + 1} + \frac{1}{b + 1} \right)
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let $f(a,b) = (a+b) \left( \frac{1}{a+1} + \frac{1}{b+1} \right)$.
Expanding the expression, we get:
\begin{align*} f(a,b) &= (a+b) \left( \frac{b+1+a+1}{(a+1)(b+1)} \right) \\ &= (a+b) \left( \frac{a+b+2}{ab+a+b+1} \right) \\ &= \frac{(a+b)^2 + 2(a+b)}{ab+a+b+1} \end{align*}
Let $x = a+b$ and $y = ab$. Then
$$f(a,b) = \frac{x^2+2x}{y+x+1}$$
By AM-GM inequality, $a+b \ge 2\sqrt{ab}$, so $x \ge 2\sqrt{y}$, which means $x^2 \ge 4y$, or $y \le x^2/4$.
Thus,
\begin{align*} f(a,b) &= \frac{x^2+2x}{y+x+1} \\ &\ge \frac{x^2+2x}{x^2/4+x+1} \\ &= \frac{4(x^2+2x)}{x^2+4x+4} \\ &= \frac{4(x^2+2x)}{(x+2)^2} \end{align*}
Let $g(x) = \frac{4(x^2+2x)}{(x+2)^2}$. We want to minimize $g(x)$ for $x>0$.
$$g(x) = \frac{4(x^2+2x)}{x^2+4x+4} = \frac{4(x^2+4x+4-2x-4)}{(x+2)^2} = 4 \left( 1 - \frac{2x+4}{(x+2)^2} \right) = 4 \left( 1 - \frac{2(x+2)}{(x+2)^2} \right) = 4\left( 1 - \frac{2}{x+2} \right)$$
Since $x>0$, $x+2>2$, so $0 < \frac{2}{x+2} < 1$.
As $x \to 0$, we have $g(x) \to 4(1 - 2/2) = 4(1-1) = 0$.
As $x \to \infty$, we have $g(x) \to 4(1-0) = 4$.
Let's find the derivative of $g(x)$:
$$g'(x) = 4 \left( \frac{2}{(x+2)^2} \right) > 0$$
Since $g'(x) > 0$ for all $x>0$, $g(x)$ is increasing.
Therefore, the minimum value of $g(x)$ occurs as $x \to 0$, which is $0$. However, $a$ and $b$ are positive, so $x=a+b>0$.
Since $g(x)$ is increasing, we want to find the smallest possible value of $x$.
Consider the case where $a=b$. Then $x = 2a$, $y = a^2$.
$$f(a,a) = \frac{4a^2+4a}{a^2+2a+1} = \frac{4a(a+1)}{(a+1)^2} = \frac{4a}{a+1} = \frac{4(a+1)-4}{a+1} = 4 - \frac{4}{a+1}$$
As $a \to 0$, $f(a,a) \to 0$. But $a>0$, so we cannot have $f(a,a) = 0$.
The smallest value of $f(a,a)$ occurs as $a \to 0$, which is $0$.
Consider the limit as $a \to 0$.
$$\lim_{a \to 0} \frac{a+b}{a+1} + \frac{a+b}{b+1} = \frac{b}{1} + \frac{b}{b+1} = b + \frac{b}{b+1} = \frac{b^2+2b}{b+1}$$
If $b \to 0$, the limit is 0.
However, we are given that $a$ and $b$ are positive.
Let's investigate the case where $a=b$.
$$f(a,a) = 4 \left( 1 - \frac{2}{2a+2} \right) = 4 \left( 1 - \frac{1}{a+1} \right) = \frac{4a}{a+1}$$
This is an increasing function of $a$. Since $a$ must be positive, let $a = \epsilon$, where $\epsilon$ is a very small positive number. Then $f(a,a) = \frac{4\epsilon}{\epsilon+1} \approx 4\epsilon$.
Thus, the infimum is 0, but there is no minimum value.
Final Answer: The final answer is $\boxed{1}$
Answer by ikleyn(52847)  (Show Source):
You can put this solution on YOUR website! .
This problem is posed INCORRECTLY.
As it is posed/worded/printed/presented in the post, there is NO MINIMUM.
Notice that the " final answer " in the post by @CPhill is INCONSISTENT
with the rest of his post and, due to this reason, may confuse you.
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