# $\frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$

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

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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

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:

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

and

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

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.

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