# Inequality regarding exponents(similar to Bernoulli's inequality)

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 $0<y<1$ 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.

• The inequality cannot be strict. At least not for the restrictions you have placed(or haven't) on a and y. For, if $a=0$ and $y \in (0,1)$ you have equality. Seems like restricting a to non-zero real values might make a better hypothesis. Jan 29, 2012 at 10:41
• I have edited the question. Jan 29, 2012 at 10:43
• 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. Jan 29, 2012 at 10:52

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.
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 $a\gt-1$, $a\ne0$.
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 $v(a)\lt v(0)+av'(0)=1+ay$.