monotonicity of $\frac{f(x)}{x}$

Question

If $f$ is defined on $[0,\infty)$, and $x_1,x_2>0$, if $\frac{f(x)}{x}$ is monotonic increasing, prove that $f(x_1+x_2)\geq f(x_1)+f(x_2)$.

I suppose it has something to do with starting from $x_1\leq x_1+x_2$ and $x_2\leq x_1+x_2$, but I can't really figure it out-- can anybody help? Thanks!

BelowAverageIntelligence · Accepted Answer · 2020-10-15T21:57:26.197

4

WLOG we may assume $x_1\geq x_2$. Then we have $$\frac{f(x_1+x_2)}{x_1+x_2}\geq \frac{f(x_1)}{x_1}$$ $$\implies f(x_1+x_2)\geq f(x_1)+x_2\frac{f(x_1)}{x_1}\geq f(x_1)+x_2\frac{f(x_2)}{x_2} = f(x_1)+f(x_2)$$which is what we wanted to show.

edited Oct 15 '20 at 21:57

answered Oct 15 '20 at 21:32

BelowAverageIntelligence

1,082

what do you mean by bellow? – BelowAverageIntelligence Oct 15 '20 at 21:35
You are upperaverageintelligence (+1). – hamam_Abdallah Oct 15 '20 at 21:37
Your answer replaces $f(x_2)$ with $f(x_1)$ without any explanation. How did you get from $\frac{x_1}{x_2}f(x_2)$ to $\frac{x_1}{x_2}f(x_1)$ – Ameet Sharma Oct 15 '20 at 21:48
sorry I believe I've fixed the problem. – BelowAverageIntelligence Oct 15 '20 at 21:54
Nice. I'd still add one more step showing the substituting in of $\frac{f(x_2)}{x_2}$ for $\frac{f(x_1)}{x_1}$ before cancelling out the $x_2$'s. – Ameet Sharma Oct 15 '20 at 21:56
is that more clear? – BelowAverageIntelligence Oct 15 '20 at 21:57
Yes. Perfect now. Thanks. – Ameet Sharma Oct 15 '20 at 21:58

monotonicity of $\frac{f(x)}{x}$

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