1965 AMC 12 Problem 34

Question

For

x\geq0,

the smallest value of

\dfrac{4x^2+8x+13}{6(1+x)}

is:

1

2

\dfrac{25}{12}

\dfrac{13}{6}

\dfrac{34}{5}

Set

t=x+1\geq1

Rewrite the expression as

\frac23t+\frac{3}{2t}

Let

t=x+1\geq1.

Since

4x^2+8x+13=4t^2+9,

the expression is

\frac23t+\frac{3}{2t}.

By AM-GM this is at least

2\sqrt{(\frac{2}{3})(\frac{3}{2})}=2,

with equality at

t=\frac{3}{2},

which is allowed.

Therefore, the correct answer is B.

Problem 34 in other years