Prove or find a counterexample $forall x>0: f(2x)-f(x)<g(3x)-g(2x)$, given information about $f, g$.












1












$begingroup$



Let $f,g:(0, infty) rightarrow mathbb{R}$ be two functions that satisfy the following for all $x>0$: $g'(x)>f'(x)$ and $f''(x)>0$.



Prove or find a counterexample:
$$
forall x>0: f(2x)-f(x)<g(3x)-g(2x).
$$




I've tried this for a long time, but I didn't make much progress.

I tried to move everything to one side and then take derivative.



If we move everything to the LHS and take derivative, we get $2f'(2x)-f'(x)+2g'(2x)-3g'(3x)$.



I don't know if this is positive or negative, and even if I knew, I don't see how it helps me. (It would help to know it if $f,g$ were defined at $0$).



Also I did not find any counterexample. Can someone please help me? And please write in your answer what was your intuition for this problem and why I still have no intuition to determine if the claim is true or false.



Thanks.






real-analysis calculus analysis functions inequality







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
    $endgroup$
    – Clayton
    Jan 21 at 20:55
















1












$begingroup$



Let $f,g:(0, infty) rightarrow mathbb{R}$ be two functions that satisfy the following for all $x>0$: $g'(x)>f'(x)$ and $f''(x)>0$.



Prove or find a counterexample:
$$
forall x>0: f(2x)-f(x)<g(3x)-g(2x).
$$




I've tried this for a long time, but I didn't make much progress.

I tried to move everything to one side and then take derivative.



If we move everything to the LHS and take derivative, we get $2f'(2x)-f'(x)+2g'(2x)-3g'(3x)$.



I don't know if this is positive or negative, and even if I knew, I don't see how it helps me. (It would help to know it if $f,g$ were defined at $0$).



Also I did not find any counterexample. Can someone please help me? And please write in your answer what was your intuition for this problem and why I still have no intuition to determine if the claim is true or false.



Thanks.






real-analysis calculus analysis functions inequality







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
    $endgroup$
    – Clayton
    Jan 21 at 20:55














1












1








1


0



$begingroup$



Let $f,g:(0, infty) rightarrow mathbb{R}$ be two functions that satisfy the following for all $x>0$: $g'(x)>f'(x)$ and $f''(x)>0$.



Prove or find a counterexample:
$$
forall x>0: f(2x)-f(x)<g(3x)-g(2x).
$$




I've tried this for a long time, but I didn't make much progress.

I tried to move everything to one side and then take derivative.



If we move everything to the LHS and take derivative, we get $2f'(2x)-f'(x)+2g'(2x)-3g'(3x)$.



I don't know if this is positive or negative, and even if I knew, I don't see how it helps me. (It would help to know it if $f,g$ were defined at $0$).



Also I did not find any counterexample. Can someone please help me? And please write in your answer what was your intuition for this problem and why I still have no intuition to determine if the claim is true or false.



Thanks.






real-analysis calculus analysis functions inequality







share|cite|improve this question











$endgroup$





Let $f,g:(0, infty) rightarrow mathbb{R}$ be two functions that satisfy the following for all $x>0$: $g'(x)>f'(x)$ and $f''(x)>0$.



Prove or find a counterexample:
$$
forall x>0: f(2x)-f(x)<g(3x)-g(2x).
$$




I've tried this for a long time, but I didn't make much progress.

I tried to move everything to one side and then take derivative.



If we move everything to the LHS and take derivative, we get $2f'(2x)-f'(x)+2g'(2x)-3g'(3x)$.



I don't know if this is positive or negative, and even if I knew, I don't see how it helps me. (It would help to know it if $f,g$ were defined at $0$).



Also I did not find any counterexample. Can someone please help me? And please write in your answer what was your intuition for this problem and why I still have no intuition to determine if the claim is true or false.



Thanks.






real-analysis calculus analysis functions inequality




real-analysis calculus analysis functions inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 21:17









Daniele Tampieri

2,3972922




2,3972922










asked Jan 21 at 20:41









OmerOmer

3619




3619








  • 2




    $begingroup$
    Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
    $endgroup$
    – Clayton
    Jan 21 at 20:55














  • 2




    $begingroup$
    Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
    $endgroup$
    – Clayton
    Jan 21 at 20:55








2




2




$begingroup$
Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
$endgroup$
– Clayton
Jan 21 at 20:55




$begingroup$
Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
$endgroup$
– Clayton
Jan 21 at 20:55










1 Answer
1






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oldest

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$begingroup$

There exists $cin (x,2x)$ such that $$f(2x)-f(x)=f'(c)x.$$



There exists $din (2x,3x)$ such that $$g(3x)-g(2x)=g'(d)x.$$



So, we need to show that $f'(c)<g'(d).$



Now, $f''(x)>0$ implies that $f'(x)$ is increasing. So we have, using that $c<d$ and $f'(x)<g'(x),$ that



$$f'(c)<f'(d)<g'(d)$$ and we are done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I like this answer. +$1$
    $endgroup$
    – Clayton
    Jan 21 at 20:56











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

There exists $cin (x,2x)$ such that $$f(2x)-f(x)=f'(c)x.$$



There exists $din (2x,3x)$ such that $$g(3x)-g(2x)=g'(d)x.$$



So, we need to show that $f'(c)<g'(d).$



Now, $f''(x)>0$ implies that $f'(x)$ is increasing. So we have, using that $c<d$ and $f'(x)<g'(x),$ that



$$f'(c)<f'(d)<g'(d)$$ and we are done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I like this answer. +$1$
    $endgroup$
    – Clayton
    Jan 21 at 20:56
















5












$begingroup$

There exists $cin (x,2x)$ such that $$f(2x)-f(x)=f'(c)x.$$



There exists $din (2x,3x)$ such that $$g(3x)-g(2x)=g'(d)x.$$



So, we need to show that $f'(c)<g'(d).$



Now, $f''(x)>0$ implies that $f'(x)$ is increasing. So we have, using that $c<d$ and $f'(x)<g'(x),$ that



$$f'(c)<f'(d)<g'(d)$$ and we are done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I like this answer. +$1$
    $endgroup$
    – Clayton
    Jan 21 at 20:56














5












5








5





$begingroup$

There exists $cin (x,2x)$ such that $$f(2x)-f(x)=f'(c)x.$$



There exists $din (2x,3x)$ such that $$g(3x)-g(2x)=g'(d)x.$$



So, we need to show that $f'(c)<g'(d).$



Now, $f''(x)>0$ implies that $f'(x)$ is increasing. So we have, using that $c<d$ and $f'(x)<g'(x),$ that



$$f'(c)<f'(d)<g'(d)$$ and we are done.






share|cite|improve this answer









$endgroup$



There exists $cin (x,2x)$ such that $$f(2x)-f(x)=f'(c)x.$$



There exists $din (2x,3x)$ such that $$g(3x)-g(2x)=g'(d)x.$$



So, we need to show that $f'(c)<g'(d).$



Now, $f''(x)>0$ implies that $f'(x)$ is increasing. So we have, using that $c<d$ and $f'(x)<g'(x),$ that



$$f'(c)<f'(d)<g'(d)$$ and we are done.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 21 at 20:50









mflmfl

26.7k12142




26.7k12142












  • $begingroup$
    I like this answer. +$1$
    $endgroup$
    – Clayton
    Jan 21 at 20:56


















  • $begingroup$
    I like this answer. +$1$
    $endgroup$
    – Clayton
    Jan 21 at 20:56
















$begingroup$
I like this answer. +$1$
$endgroup$
– Clayton
Jan 21 at 20:56




$begingroup$
I like this answer. +$1$
$endgroup$
– Clayton
Jan 21 at 20:56


















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