Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $a,b,c,d >0$ and $a+b+c+d=2$. Prove this: $$ \frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$$

share|cite|improve this question
Remark: $16/7$ is what you get when $a=b=c=d=1/2$. – 1015 Jan 2 '13 at 15:45
That value is also assumed at $a=0$, $b=c=d=2/3$. Essentially, $1/(3x^2+1)$ is concave in $[1/3,2]$ so if $a,b,c,d\geq 1/3$ it is minimized when $a=b=c=d$, but then you have to deal with the case where some are in $[0,1/3]$ – Thomas Andrews Jan 2 '13 at 17:02
It's pretty easy to show if $a,b,c\in[0,1/3]$ and $d=2-a-b-c$ then $f(a)+f(b)+f(c)+f(d)\geq 3f(1/3)+f(2) > 16/7$. So we can reduce to the case where at most $2$ of $a,b,c,d$ are in $[0,1/3)$ and the values outside that interval are equal (again by concavity.) – Thomas Andrews Jan 2 '13 at 17:13
I think it's can be solved by Cauchy Schwarz by nice solution but i can't find it now – Haruboy15 Jan 3 '13 at 15:13
up vote 3 down vote accepted

This is a brute force approach.

First, let's show it for $a,b,c,d\geq 0$, since it is true in that case, too, and this set is compact, so if there is a minimum, it is reached somwhere.

Let $f(x)=\frac{1}{3x^2+1}$. Then $f''(x)=\frac{6(9x^2-1)}{(3x^2+1)^3}$. So $f(x)$ is convex on $[1/3,2]$. In particular, if $a,b,c,d\geq 1/3$ then $$f(a)+f(b)+f(c)+f(d)\geq 4f\left(\frac{a+b+c+d}4\right)=4f\left(\frac 1 2\right) = \frac{16}7$$

So, to find a counter-example, we need some of $a,b,c,d$ to be in $[0,1/3)$. For now, assume $a\leq b\leq c\leq d$. By convexity of $f(x)$, we can assume that any values $\geq 1/3$ are equal.

It can be shown pretty directly that if $a,b,c<1/3$ and $d=2-(a+b+c)$ that: $$f(a)+f(b)+f(c)+f(d)\geq f(1/3)+f(1/3)+f(1/3)+f(2) > 16/7$$

So, if $f(a)+f(b)+f(c)+f(d)$ takes any value smaller than $16/7$, it must be with:

$$a<1/3, c=d\geq 1/3$$

Now, if $a,b\in (0,1/3)$, then consider $f(a-\delta)+f(b+\delta)$ for small $\delta>0$. By the intermediate value theorem, $f(a-\delta) = f(a) - f'(a_0)\delta$ for some $a-\delta<a_0<a$ and $f(b+\delta) = f(b)+f'(b_0)\delta$ for some $b<b_0<b+\delta$. So $$f(a-\delta) + f(b+\delta) = f(a)+f(b) + \delta(f'(b_0)-f'(a_0))$$ Since $f''$ is negative on $(0,1/3)$, and $b_0>a_0$, then $f'(b_0)<f'(a_0)$, so we get:

$$f(a+\delta)+f(b-\delta) < f(a)+f(b)$$

So if there is a minimum reached with $0\leq a\leq b\leq 1/3$, then that minimum must be reached with either $a=0$ or $b=1/3$. But once $b\in[1/3,2]$, it is in the region of convexity, so we can get a miminum with $a\in[0,1/3)$ and $b=c=d$.

So we've reduced the cases to:

$$a=0\leq b < 1/3 <c=d=1-\frac{b}{2}$$


$$0\leq a < 1/3 < b=c=d=\frac{2-a}{3}$$

Note the minimum is actually reached when $a=0$ and $b=c=d=2/3$, so we have to take care with each of these cases. We essentially need to minimize the two formulas:

$$1+f(b) + 2f\left(1-\frac{b}{2}\right), 0\leq b<1/3$$

and $$f(a) + 3f\left(\frac{2-a}{3}\right), 0\leq a< 1/3$$

The fact that the first is minimized when $b=1/3$ and the second when $a=0$ are not obvious to me, but that is what Wolfram Alpha says. We compute from there see we get at least $16/7$ in both cases.

Given that the region $a,b,c,d\geq 0$ and $a+b+c+d=2$ is a regular tetrahedron, and the minimum value is assumed at the center of mass and at the centers of the faces, it seems like you might be able to make a geometric argument, rather than this brute force approach. I'm just not seeing it.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.