Prove that $y=e^x$ always lies above any tangent drawn on it.

Question

I want to prove that all of the graph of $y=e^x$ lies above the tangent of $y=e^x$ at the point $x=a$, for all real $a$. (The graph of $y=e^x$ lies above its tangent except at $x=a$, where the graph touches the tangent).

By differentiating, I get that the equation of the tangent of $y=e^x$ at $x=a$ is $y=e^ax+e^a-ae^a$.

If I can show that $e^a(x+1-a)$ $\leq$ $e^x$, for all $a$, with equality only when $x=a$, then I'm done.

How can I prove this inequality? Or, is there a better way to go about proving this?

Answer 1

Astonishingly much about the exponential function (cf. most of my answers to questions about the exponential function) can be shown using virtually only its multiplicativity $$\tag1 e^{x+y}=e^xe^y $$ and the lower bound $$\tag2 e^x\ge 1+x\qquad\text{with equality iff }x=0.$$ Same here: Extract a (positive) factor $e^a$ to arrive at the desired claim: $$ e^x=e^{x-a}e^a\ge (1+x-a)e^a\qquad\text{with equality iff }x=a.$$

Of course this is cheating as $(2)$ is in fact the special case of the claim for $a=0$, so we better actually prove it this time before using it.

Recall that the exponential is always positive, hence has positive derivative, hence is strictly increasing. First consider $x>0$. By the MWT, there is some $\xi\in(0,x)$ with $$\frac{e^x-e^0}{x-0}=e^\xi>e^0,$$ so $$ e^x>x+1$$ as desired. Next consider $x<0$. This time, there is some $\xi\in(x,0)$ with $$\frac{e^x-e^0}{x-0}=e^\xi<e^0,$$ so again (careful with the sign!) $$ e^x>x+1$$ as desired.

Answer 2

Take the derivative of $$D(x)=e^a(x+1-a)-e^x$$ $$D'=e^a-e^x$$ Then $D'=0$ means $x=a$ and $D=0$. This means that the exponential is always on one side of the tangent, except at the point of tangency. Then, to prove that it's above, all you need is one point. I would choose the limit at $-\infty$. Abusing the notation a little $D(-\infty)=-\infty-0=-\infty$, so the tangent is below the exponential.

Answer 3

$e^{y} \geq 1+y$ for all real numbers $y$. Put $y=x-a$ to get $e^{x} e^{-a} \geq 1+x-a$ or $e^{x} \geq e^{a}(1+x-a)$. Also equality holds in $e^{y} \geq 1+y$ iff $y=0$ so equality holds in $e^{x} \geq e^{a}(1+x-a)$ iff $x=a$.

Proof of the inequality $e^{y} \geq 1+y$: Let $f(y)=e^{y}-1-y$. Then $f(0)=0$ and $f'(y)=e^{y} -1$ which is positive for $y>0$ and neagative for $y<0$. Hence $f$ is strictly increasing in $(0,\infty)$ and strictly decreasing in $(-\infty, 0)$. This implies that $f(y) \geq 0$ and $f(y)=0$ iff $y=0$.