Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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.
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 $v(a)\lt v(0)+av'(0)=1+ay$.
