0
$\begingroup$

I was trying to solve a problem today when I felt that if the following is proved, we are done:

$(1+a)^y<(1+ay)$ for $0 for all non-zero reals a,y.I am not able to make much progress on this one.Can anyone please help me on this?

As always,thank you.

  • 1
    If you're familiar with the version of Bernouli's inequality that says 1+rx<(1+x)^r when r>1, x>-1, $x\neq 0$, then this follows from setting $rx=a$, $y=\frac{1}{r}$, and taking the $y$ power of both sides.2012-01-29

1 Answers 1

1

First method: convexity.

The function $v:a\mapsto(1+a)^y$ has second derivative v''(a)=y(y-1)(1+a)^{y-2}\lt0 at $a\gt-1$ hence $v$ is concave on $a\geqslant-1$. Since v'(0)=y, $t:a\mapsto1+ay$ is the equation of its tangent at $a=0$. The graph of a concave function lies below each of its tangents hence $v(a)\lt t(a)$ at every $a\gt-1$ except $a=0$ where $v(0)=t(0)$ and we are done.

Second method: logarithms.

Let $u(a)=\log(1+ay)-y\log(1+a)$. The assertion to prove is equivalent to $u(a)\gt0$. Note that $u(0)=0$ and u'(a)=\frac{y}{1+ay}-\frac{y}{1+a}=\frac{ay(1-y)}{(1+ay)(1+a)} is positive on $a\gt0$ and negative on $a\lt0$. Thus, the function $u:a\mapsto u(a)$ decreases to $u(0)=0$ on $a\lt0$, then increases from $u(0)=0$ on $a\gt0$. In particular, $u(0)=0$ and $u(a)\gt0$ for every $a\gt-1$, $a\ne0$.

Third method: Taylor formula.

Consider once again the function $v:a\mapsto(1+a)^y$. Then v(a)=v(0)+av'(0)+\frac12a^2v''(x_a) for some $x_a$ between $0$ and $a$, by Taylor formula. Now, v''(x)=y(y-1)(1+x)^{y-2}\lt0 for every $x\gt-1$, hence v''(x_a)\lt0. Since $v(0)=1$, v'(0)=y, this leaves us with the inequality v(a)\lt v(0)+av'(0)=1+ay.