Prove that if $f(x)=a/(x+b)$ then $f((x_1+x_2)/2)\le(f(x_1)+f(x_2))/2$

Question

This exercise :

If $f(x)=a/(x+b)$ then : $$ f((x_1+x_2)/2)\le(f(x_1)+f(x_2))/2$$

was in my math olympiad today (for 16 years olds). I proved this by saying this is true due to Jensen's inequality.

Is this an acceptable answer?(with leaving aside the fact that i didn't prove the function is convex since it can't be done with the info we got until now). Does it have any other way of proving this?

In questions like this, you should be more careful with your statement of the problem. What are the allowed values of $x$, $a$, and $b$? Obviously there must be restrictions $-$ for instance, if $x = -b$, then $f(x)$ is not defined; and if $a < 0$, then the statement is false. — TonyK, Jan 09 '16 at 18:32
No, you have to do better than that. You don't say whether $x$ is allowed to be equal to $b$; and you don't give any restrictions on $a$. Surely the exam question had these details? — TonyK, Jan 09 '16 at 18:39

score 2 · Answer 1 · answered Jan 09 '16 at 17:54

2

I'll use $x$ and $y$ instead of $x_1,x_2$ .

You can just calculate and see what you get :

$$\frac{1}{2} \cdot \left (\frac{a}{x+b}+\frac{a}{y+b} \right )>\frac{2a}{2b+x+y}$$

$$\frac{1}{x+b}+\frac{1}{y+b} \geq \frac{4}{x+y+2b}$$

You can prove this by a simple calculation :

$$\frac{(x-y)^2}{(x+b)(y+b)(x+y+2b)} \geq 0$$ which is obvious .

You can also prove it by Cauchy-Schwarz in a special form :

$$\frac{1}{x+b}+\frac{1}{y+b} \geq \frac{(1+1)^2}{x+y+2b}=\frac{4}{x+y+2b}$$

answered Jan 09 '16 at 17:54

is it wrong to say this is true due to jensen's inequality? – Sierra Res Jan 09 '16 at 17:55
@KristianMihalaj It isn't but you must prove that $f$ is convex . – Jan 09 '16 at 17:56

score 1 · Answer 2 · answered Jan 09 '16 at 18:20

1

You can prove that f is convex by finding it second derivative, which is positive.

answered Jan 09 '16 at 18:20

Imawesome

295

Prove that if $f(x)=a/(x+b)$ then $f((x_1+x_2)/2)\le(f(x_1)+f(x_2))/2$

2 Answers2