An inequality on a convex function An exercise in my textbook asked to prove the following inequality, valid for all $a,b,c,d \in R $
$$\left(\frac{a}{2} + \frac{b}{3} + \frac{c}{12} + \frac{d}{12}\right)^4 \leq \frac{a^4}{2} + \frac{b^4}{3} + \frac{c^4}{12} + \frac{d^4}{12}$$
There is a straightforward proof using Convex Functions:


*

*$f(x) = x^4$ is a convex function satisfying $f(\lambda x + (1-\lambda) y) \leq \lambda f(x) + (1 - \lambda)f(y)$ for all $x,y \in R$ and $\lambda \in [0,1]$

*Since $\frac{1}{2} + \frac{1}{3} + \frac{1}{12} + \frac{1}{12} = 1$, we can use the convexity property to obtain the inequality.


Since this question was on the chapter about Convex Functions, I was able to find the solution quickly. However, had I seen the problem in a "standalone" manner I would probably take longer to solve it, and at least spend a lot of muscle opening up the left hand term :)
My question is: What would be other ways to obtain this same result? What if someone had shown me this problem back when I was in eight grade?

By Holder 
$$\frac{a^4}{2} + \frac{b^4}{3} + \frac{c^4}{12} + \frac{d^4}{12}=\left(\frac{a^4}{2} + \frac{b^4}{3} + \frac{c^4}{12} + \frac{d^4}{12}\right)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{12} + \frac{1}{12}\right)^3\geq$$
$$\geq\left(\frac{|a|}{2} + \frac{|b|}{3} + \frac{|c|}{12} + \frac{|d|}{12}\right)^4\geq\left(\frac{a}{2} + \frac{b}{3} + \frac{c}{12} + \frac{d}{12}\right)^4$$
and we are done!
