3
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ What about $(e^{x/2}-e^{y/2})^2$? $\endgroup$ – Sorfosh Dec 24 '18 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 '18 at 18:30
2
$\begingroup$

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 $x+y+z=\pi.$

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

Rephrasing:

$e^x+e^y=$

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

$2e^{x/2}e^{y/2};$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Plotting is not a proof. It may be suggestive, as in this case, but the suggestion needs to be made rigorous. $\endgroup$ – marty cohen Dec 24 '18 at 19:17
  • 1
    $\begingroup$ Because The second dericvative of The exponencial function is always positive, The epigraph is convex. $\endgroup$ – Tito Eliatron Dec 24 '18 at 20:54
  • $\begingroup$ Now that is a proof. $\endgroup$ – marty cohen Dec 24 '18 at 21:01
1
$\begingroup$

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 $$2f(z)<f(x)+f(y)+(2z-x-y)f'(z)=f(x)+f(y).$$

$\endgroup$
1
  • $\begingroup$ By the same method, if $r\in (0,1)$ then $rf(x)+(1-r)f(y)>f(rx+(1-r)y)$. $\endgroup$ – DanielWainfleet Dec 25 '18 at 0:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.