# Is this a right step for inequalities? (Proof needed if it is right)

I know that if $a,b,c,d$ are positive numbers then we have $$\frac{a+c}{b+d}\leq \max\{\frac{a}{b},\frac{c}{d}\}$$ Can I use this to write as follows $$\frac{f_1+f_2+f_3+\cdots f_K}{g_1+g_2+g_3+\cdots g_K+Kc}\leq \max\{\frac{f_1}{g_1+c},\frac{f_2}{g_2+c},\cdots \frac{f_K}{g_K+c}\}$$ where $f_i$'s and $g_i$'s are positive numbers and $c$ is also a positive number. If this is right then how to formally prove it? Any help in this regard will be much appreciated. Thanks in advance.

• Apr 2 '18 at 7:06

For $K=3$, \begin{align*} \dfrac{f_{1}+f_{2}+f_{3}}{g_{1}+g_{2}+g_{3}+3c}&=\dfrac{f_{1}+(f_{2}+f_{3})}{(g_{1}+c)+((g_{2}+c)+(g_{3}+c))}\\ &\leq\max\left\{\dfrac{f_{1}}{g_{1}+c},\dfrac{f_{2}+f_{3}}{(g_{2}+c)+(g_{3}+c)}\right\}\\ &\leq\max\left\{\dfrac{f_{1}}{g_{1}+c},\max\left\{\dfrac{f_{2}}{g_{2}+c},\dfrac{f_{3}}{g_{3}+c}\right\}\right\}\\ &=\max\left\{\dfrac{f_{1}}{g_{1}+c},\dfrac{f_{2}}{g_{2}+c},\dfrac{f_{3}}{g_{3}+c}\right\}. \end{align*}