The inequality is $$\frac{e^x+e^y}{2}>e^\frac{x+y}{2}$$ where $x \neq y$.

This is my first time coming across inequalities of this form thus I really don't know to approach it correctly. Here is what I tried:

Let us define $f(p)=e^p>0$. Then $f'(p)=e^p>0$ and $f''(p)=e^p>0$ for all $p$. Thus the function is real valued and convex? (*)

Therefore, by Jensen's inequality, we have that

$$e^\frac{x+y}{2}=f\left(\frac{1}{2}x+\frac{1}{2}{y}\right) \leq \frac{1}{2}f(x)+\frac{1}{2}f(y)=\frac{e^x+e^y}{2}$$

Since $f''(p)=e^p>0$ is always a strict inequality, then I was thinking this makes the inequality also strict, so that we have

$$e^\frac{x+y}{2}=f\left(\frac{1}{2}x+\frac{1}{2}{y}\right) \lt \frac{1}{2}f(x)+\frac{1}{2}f(y)=\frac{e^x+e^y}{2}$$

My question, (besides asking whether or not this 'proof' is correct), is what is the general strategy of proving such inequalities? And is this a correct application of the Jensen inequality? Was it right to have had concluded that the function is convex at (*)? And a request for a hint to the right answer, in the case that this one is completely wrong. Any feedback is highly appreciated!

EDIT: I appreciate the answers coming in about using the AM-GM inequality, but I'd also like to know if this proof presented here works too.

  • 3
    $\begingroup$ What about $(e^{x/2}-e^{y/2})^2$? $\endgroup$
    – Sorfosh
    Dec 24, 2018 at 18:29
  • 3
    $\begingroup$ You could use the arithmetic-geometric mean inequality to prove your statement directly. It’s usually written as $\frac{x+y}{2} \geq \sqrt{xy}$ with equality if and only if $x = y$. $\endgroup$
    – user328442
    Dec 24, 2018 at 18:30

5 Answers 5


Use AM-GM inequality!

$$\frac{e^x+e^y}{2}\geq e^\frac{x+y}{2}$$

Equality holds only when $e^x=e^y$ or $x=y$, which isn't the case.




$(e^{x/2}-e^{y/2})^2 +2e^{x/2}e^{y/2} \ge$


Equality for $e^{x/2}=e^{y/2}$.


Your proof is right and the function is indeed convex.

But the inequality like Jensen can be true even the function is not convex.

For example, the function $f(x)=-\cos{x}$ is not convex function on $[0,\pi],$

but the following inequality is true: $$\frac{-\cos{x}-\cos{y}-\cos{z}}{3}\geq-\cos\frac{x+y+z}{3}$$ for all $\{x,y,z\}\subset[0,\pi]$ such that $x+y+z=\pi.$


A simpler proof.

Assume $x<y$.

Plot $f(x)=e^x$. The arc of curve Jon int $(x,f(x))$ and $(y,f(y))$. Plot The segment joining these two points.

The segment is above The curve.

The midpoint of The segment is above The midpoint of The curve.

The midpoint of The segment has y coordínate $(e^x+e^y)/2$ and The midpoint of The curve, $e^{(x+y)/2}$.

We are done

  • $\begingroup$ Plotting is not a proof. It may be suggestive, as in this case, but the suggestion needs to be made rigorous. $\endgroup$ Dec 24, 2018 at 19:17
  • 1
    $\begingroup$ Because The second dericvative of The exponencial function is always positive, The epigraph is convex. $\endgroup$ Dec 24, 2018 at 20:54
  • $\begingroup$ Now that is a proof. $\endgroup$ Dec 24, 2018 at 21:01

Suppose $f''>0.$ Let $x<y$ and let $z=(x+y)/2.$

Method 1.There exists $a\in (x,z)$ with $$(1).\quad f(x)=f(z)+(x-z)f'(z)+(x-z)^2f''(a)/2.$$ There exists $b\in (z,y)$ with $$(2).\quad f(y)=f(z)+(y-z)f'(z)+(y-z)^2f''(b)/2.$$ Adding (1) and (2), since $x+y-2z=0,$ we have $$f(x)+f(y)=2f(z)+(x+y-2z)f'(z)+(x-z)^2f''(a)/2+(y-z)^2f''(b)/2=$$ $$=2f(z)+(x-z)^2f''(a)/2+(y-z)^2f''(b)/2>$$ $$>2f(z).$$

Method 2. $f'$ is strictly increasing and continuous so we have $$(1'). \quad f(z)=f(x)+\int_x^zf'(t)dt<f(x)+(z-x)f'(z)$$ and we have $$(2').\quad f(z)=f(y)+\int_y^zf'(t)dt<f(y)+(z-y)f'(z).$$ Adding (1') and (2'), since $2z-x-y=0, $ we have $$2f(z)<f(x)+f(y)+(2z-x-y)f'(z)=f(x)+f(y).$$

  • $\begingroup$ By the same method, if $r\in (0,1)$ then $rf(x)+(1-r)f(y)>f(rx+(1-r)y)$. $\endgroup$ Dec 25, 2018 at 0:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .