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Let $m = \min \left\{ a, \frac{1}{b}, b^2 + \frac{1}{a+1} \right\}$.
\n" ); document.write( "We want to find the largest possible value of $m$.
\n" ); document.write( "Since $m$ is the minimum of the three quantities, we have
\n" ); document.write( "$$ m \le a, \quad m \le \frac{1}{b}, \quad m \le b^2 + \frac{1}{a+1}. $$
\n" ); document.write( "We want to maximize $m$, so we consider the case when all three quantities are equal to $m$:
\n" ); document.write( "$$ a = m, \quad \frac{1}{b} = m, \quad b^2 + \frac{1}{a+1} = m. $$
\n" ); document.write( "From $a=m$ and $\frac{1}{b} = m$, we get $a=m$ and $b=\frac{1}{m}$.
\n" ); document.write( "Substitute $a=m$ and $b=\frac{1}{m}$ into the third equation:
\n" ); document.write( "$$ \left( \frac{1}{m} \right)^2 + \frac{1}{m+1} = m. $$
\n" ); document.write( "$$ \frac{1}{m^2} + \frac{1}{m+1} = m. $$
\n" ); document.write( "Multiplying by $m^2(m+1)$, we get
\n" ); document.write( "$$ m+1 + m^2 = m^3(m+1). $$
\n" ); document.write( "$$ m+1 + m^2 = m^4 + m^3. $$
\n" ); document.write( "$$ m^4 + m^3 - m^2 - m - 1 = 0. $$
\n" ); document.write( "Let $f(m) = m^4 + m^3 - m^2 - m - 1$. We are looking for a positive real root.
\n" ); document.write( "We can test some values:
\n" ); document.write( "$f(1) = 1+1-1-1-1 = -1$
\n" ); document.write( "$f(2) = 16+8-4-2-1 = 17$
\n" ); document.write( "Since $f(1) < 0$ and $f(2) > 0$, there is a root between 1 and 2.
\n" ); document.write( "Let's consider the case when the three quantities are close to each other.
\n" ); document.write( "If we set $a = \frac{1}{b}$, then $ab = 1$.
\n" ); document.write( "Then we have $a = m$, $b = \frac{1}{m}$, and $b^2 + \frac{1}{a+1} = m$.
\n" ); document.write( "$$ \frac{1}{m^2} + \frac{1}{m+1} = m. $$
\n" ); document.write( "Let us try to find a simpler solution.
\n" ); document.write( "Suppose $a = \frac{1}{b} = b^2 + \frac{1}{a+1} = m$.
\n" ); document.write( "If $a = \frac{1}{b}$, then $ab=1$.
\n" ); document.write( "If $a = m$ and $\frac{1}{b} = m$, then $b=\frac{1}{m}$.
\n" ); document.write( "If $m = b^2 + \frac{1}{a+1}$, then $m = \frac{1}{m^2} + \frac{1}{m+1}$.
\n" ); document.write( "If we assume $m = 1$, then $a = 1$, $b = 1$, and $1 = 1^2 + \frac{1}{1+1} = 1 + \frac{1}{2} = \frac{3}{2}$, which is false.
\n" ); document.write( "Let's try to find a solution by setting $a = \frac{1}{b}$.
\n" ); document.write( "Then $m \le a$ and $m \le \frac{1}{b}$, so $m \le a$ and $m \le a$.
\n" ); document.write( "Also, $m \le b^2 + \frac{1}{a+1}$.
\n" ); document.write( "Since $b = \frac{1}{a}$, $m \le \frac{1}{a^2} + \frac{1}{a+1}$.
\n" ); document.write( "If $m = a$, then $a \le \frac{1}{a^2} + \frac{1}{a+1}$.
\n" ); document.write( "$a^3(a+1) \le a+1 + a^2$
\n" ); document.write( "$a^4 + a^3 \le a^2 + a + 1$
\n" ); document.write( "Let's consider the case when $a = \frac{1}{b} = b^2 + \frac{1}{a+1}$.
\n" ); document.write( "If $a=1$, then $b=1$.
\n" ); document.write( "$1 = 1^2 + \frac{1}{1+1} = 1 + \frac{1}{2} = \frac{3}{2}$, which is false.
\n" ); document.write( "If $m=1$, then $a=1$ and $b=1$.
\n" ); document.write( "Then $m = \min \left\{ 1, 1, 1^2 + \frac{1}{1+1} \right\} = \min \left\{ 1, 1, \frac{3}{2} \right\} = 1$.
\n" ); document.write( "Thus, $m=1$ is a possible value.
\n" ); document.write( "Let $a = 1.32$. $b = 1/a = 0.7575$. $b^2 + 1/(a+1) = 0.7575^2 + 1/2.32 = 0.5738 + 0.4310 = 1.0048$
\n" ); document.write( "If $m=1$, then $1 \le a$, $1 \le 1/b$, $1 \le b^2 + 1/(a+1)$.
\n" ); document.write( "$b \le 1$, $b^2 \ge 1-1/(a+1)$
\n" ); document.write( "$b^2 \ge a/(a+1)$.
\n" ); document.write( "$1/b \ge 1$, so $1/b \ge a$.
\n" ); document.write( "$1 \ge a/(a+1)b^2$, so $1/b^2 \ge a/(a+1)$.
\n" ); document.write( "$1/b^2 = a^2 \ge a/(a+1)$.
\n" ); document.write( "$a^2(a+1) \ge a$.
\n" ); document.write( "$a(a+1) \ge 1$.
\n" ); document.write( "$a^2+a-1 \ge 0$.
\n" ); document.write( "$a = (-1 \pm \sqrt{1+4})/2 = (-1 \pm \sqrt{5})/2$.
\n" ); document.write( "$a \ge (-1 + \sqrt{5})/2$.
\n" ); document.write( "If $m=1$, then $a \ge 1, b \le 1$, and $b^2 + \frac{1}{a+1} \ge 1$.
\n" ); document.write( "If $a=1$, $b=1$, $1 \le 1, 1 \le 1, 1 \le 3/2$.
\n" ); document.write( "Thus $m=1$ is possible.
\n" ); document.write( "Final Answer: The final answer is $\boxed{1}$
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