My Habit of Overcomplicating things
by SomeonecoolLovesMaths, Nov 17, 2024, 7:28 AM
I always somehow tend to overcomplicate simple things. Here are just a few instances of my providing overkill solutions to problems and then crying about it for hours. 
These are some of the recent problems that I have overkilled. I might add more to this list later, if I remember ofcourse because I guess I am too consistent with my blog.
be positive real numbers such that
Prove that 
.![$$1- \frac{a}{1+a}+ 1 -\frac{b}{1+b}+1 -\frac{c}{1+c} = 3-1 = \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \Longrightarrow \frac{\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \right)}{3} = \frac{2}{3} \geq \frac{3}{2a+2b+2c+3} \Longrightarrow 2a+2b+2c \geq 3 \Longrightarrow \frac{2a+2b+2c}{3} = 1 \geq \sqrt[3]{8abc} \Longrightarrow abc \leq \frac{1}{8}$$](http://latex.artofproblemsolving.com/8/4/c/84c8ba18fb8246a1d89f24fb9476dadd508c105e.png)
.
![$$\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b} = \frac{a}{a\cdot a^2+b \cdot b+c\cdot 1}+\frac{b}{b\cdot b^2+c \cdot c+a\cdot 1}+\frac{c}{c\cdot c^2+a \cdot a+b\cdot 1} = \frac{1}{3} \left( \frac{a}{\frac{a\cdot a^2+b \cdot b+c\cdot 1}{a+b+c}}+\frac{b}{\frac{b\cdot b^2+c \cdot c+a\cdot 1}{a+b+c}}+\frac{c}{\frac{c\cdot c^2+a \cdot a+b\cdot 1}{a+b+c}} \right) \leq \frac{1}{3} \left( \frac{a}{\sqrt[3]{a^2b}} + \frac{b}{\sqrt[3]{b^2c}} + \frac{c}{\sqrt[3]{c^2a}} \right) = \frac{1}{3} \left( \sqrt[3]{\frac{a}{b}} + \sqrt[3]{\frac{b}{c}} + \sqrt[3]{\frac{c}{a}} \right) \leq \frac{1}{3} \left( (a+b+c)^{\frac{1}{3}} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)^{\frac{1}{3}} (3)^{\frac{1}{3}} \right) \leq \frac{1}{3} (27)^{\frac{1}{3}} = 1$$](http://latex.artofproblemsolving.com/1/4/d/14d5888d3629e93cbd2b102cfdec759c7d1cc1a9.png)
be a positive real numbers such that 
.
These are some of the recent problems that I have overkilled. I might add more to this list later, if I remember ofcourse because I guess I am too consistent with my blog.
March 2025
February 2025