# 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?

It all begins with the Cauchy-Schwarz inequality which says that the average of the squares is bigger than the square of the average: $$\left({x+y\over 2}\right)^2\leq {x^2+y^2\over 2}.$$ We see this is true since subtracting the left from the right hand expression gives $(x-y)^2/4\geq 0$.

Next, we extend this result to collections larger than two:
$$\left({x_1+x_2+\cdots +x_n\over n}\right)^2\leq {x_1^2+x_2^2+\cdots+x_n^2\over n}.$$ You can prove this by induction on $n$, or directly by subtracting the left from the right hand expression to get the obviously non-negative $\sum_{i<j} (x_i-x_j)^2/n^2$.

Now, applying this inequality twice, once to the numbers $x_j$ and then to the numbers $x_j^2$, gives the fourth power version: $$\left({x_1+x_2+\cdots+x_n\over n}\right)^4\leq \left({x_1^2+x_2^2+\cdots+x_n^2\over n}\right)^2 \leq {x_1^4+x_2^4+\cdots+x_n^4\over n}.$$

Finally, as mentioned by Carl, putting $n=12$ and choosing $x_j=a$ for $j=1,2,3,4,5,6$, $x_j=b$ for $j=7,8,9,10$, $x_{11}=c$, and $x_{12}=d$ then plugging into the formula above gives your result.

I don't know how well this would go over in a grade 8 classroom, but it doesn't use any advanced mathematics.

• Nice presentation, including the trick of iterating Cauchy-Schwarz to prove the degree $4$ case. It happens that one can prove directly $((x+y)/2)^4\le(x^4+y^4)/2$ modifying the trick you used for $n=2$ and squares: the difference between the right and the left hand expression is $(x-y)^2(7x^2+10xy+7y^2)/16$ and the last polynomial has no real root. Maybe one could make the trick work directly in the $n$ case as well but right now I fail to see how. – Did Apr 16 '11 at 5:45

Let $\{a_j\}$ be a set of $n$ numbers. Then the 4th power of the average of $\{a_j\}$ is smaller than the average of the 4th powers of $\{a_j\}$: $$(\Sigma_ja_j/n)^4 \le \Sigma_j a_j^4/n.$$ With $n=12$ and appropriate choice of $a_j$ this will give you what you want.

Unfortunately, being an old foggy, I can't recall why it is that I know the above fact, but it should be easy to show with calculus.

• Ah, here it is, Hölder's inequality: en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality , that is, written with the expectations of elementary probability theory, $E\bigl[|X|^r\bigr]\le\bigl(E\bigl[|X|^s\bigr]\bigr)^{r/s}$ with $r=s=4$. Hmmmmm. Maybe that's not quite 8th grade. – Carl Brannen Apr 15 '11 at 22:55
• You mean $r=1$ here. – user940 Apr 16 '11 at 2:42

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!