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Let $a,b,c,d\ge 0$,and such $a+b+c+d=1$, show that $$3(a^2+b^2+c^2+d^2)+64abcd\ge 1$$

use AM-GM $$a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$$ it suffices to $$4\sqrt{abcd}+64abcd\ge 1$$

Martin Sleziak
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math110
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    Consider $a=1, b=c=d=0$, your approach would fail :( – Macavity Mar 15 '16 at 07:24
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    Also there is equality when all variables are equal or with a permutation of $a=0, b=c=d=\frac13$, so cant be straightforward AM-GM! – Macavity Mar 15 '16 at 07:36
  • I know my approach is fail,Thank you. But also $a=b=c=d=\dfrac{1}{4}$ is equal – math110 Mar 15 '16 at 07:38
  • So some form of Schur may be involved... – Macavity Mar 15 '16 at 07:42
  • I smell Newton's inequalities. – orion Mar 15 '16 at 08:37
  • This issue looks connected with the concepts presented in the interesting article "One Dimensional Metrical Geometry" of N.J. Wildberger http://arxiv.org/pdf/math/0701338 Not only this would probably yield another proof of this problem, but of other problems of the same kind. – Jean Marie Mar 15 '16 at 10:10

4 Answers4

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Here is a way without Schur. Let $p = a+b, u = ab$ and $q = c+d, v = cd$. Then we have $p+q = 1$ and wish to show $$F = 3(p^2+q^2-2u-2v)+64uv \ge 1$$

For any fixed $p, q$, the variables $u, v$ can range continuously in $u \in [0, p^2/4]$ and $v \in [0, q^2/4]$. Since $F$, viewed as a function of one variable (either $u$ or $v$) is linear, the minimum has to be when $u, v$ take one of the extreme points.

Now if $u = p^2/4$ and $v = q^2/4$, the variables are equal and we have $F \ge 1 \iff (p-q)^2(p^2+q^2)\ge 0$ so we have equality when $p=q \implies a=b=c=d$ in this case.

Else we have $uv=0$, so WLOG let $d=0$ and the inequality is now the obvious $$a+b+c=1 \implies 3(a^2+b^2+c^2) \ge 1$$

Macavity
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Here is a way using Schur and smoothing. Let $a$ be the minimum among $a, b, c, d$ and say $3p = b+c+d, \,q = bc+cd+db, \,r = bcd$. We define $$F(a, b, c, d) = 3\sum_{cyc} a^2+64abcd $$

We will first show $F(a, b, c, d) \ge F(a, p, p, p)$. $$\iff 3(b^2+c^2+d^2-3p^2) \ge 64a(p^3-r)$$ But $b^2+c^2+d^2-3p^2 = 2(3p^2-q)$, so we have to show $$\iff 3(3p^2-q) \ge 32a(p^3-r) \tag{1}$$ By Schur inequality $$\sum_{cyc} b^3+3bcd \ge \sum_{cyc} bc (b+c) \implies \frac43p(3p^2-q) \ge p^3-r$$ so it is enough for $(1)$ to show $$3(3p^2-q) \ge 32a\cdot \frac43p(3p^2-q) \iff (3p^2-q)(9-128ap) \ge 0 $$

Now $3p^2 \ge q$ is easily shown and $9-128ap = 9-128(1-3p)p > 0 $ as $p \ge \frac14$ for $a$ to be the minimum. Hence we have shown $(1)$ holds true.

Now all that is needed is to show $F(1-3p, p, p, p) \ge 1$. $$\iff 3((1-3p)^2+3p^2)+64(1-3p)p^3 \ge 1$$ $$\iff 2(1+2p)(1-3p)(1-4p)^2 \ge 0$$ which is now obvious. Equality is iff $3p=1$, i.e. $a=0, b=c=d=\frac13$ or when $4p=1$, i.e. $a=b=c=d=\frac14$.

Macavity
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the inequality equals:

$3(a+b+c+d)^2(a^2+b^2+c^2+d^2)+64abcd-(a+b+c+d)^4\ge 0 $

WLOG, let $d=min(a,b,c,d)$

$3(a+b+c+d)^2(a^2+b^2+c^2+d^2)+64abcd-(a+b+c+d)^4=2d\left(\dfrac{1}{3}\sum_{cyc (a,b,c)}(a-d)^2(2a+d)+\sum_{cyc (a,b,c)} a(a-d)^2+(a+b+c-3d)((a-b)^2+(b-c)^2+(a-c)^2)+3(a^3+b^3+c^3+3abc-\sum ab(a+b))+\dfrac{a^3+b^3+c^3}{3}-abc\right)+(a+b+c-3d)^2((a-b)^2+(b-c)^2+(a-c)^2)\ge0$

chenbai
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Maybe a generalization (not necessarily an easy way toward a solution):

Let $M_k=\sqrt[k]{(a^k+b^k+c^k+d^k)/4}$ be a $k$-th mean of the numbers, and $M_0=\sqrt[4]{abcd}$ the limit special case (geometric mean). Note that $M_1=1/4$. The question then is $$3\cdot 4 M_2^2+64M_0^4\geq 1$$ $$\frac{3 M_2^2+M_0^4/M_1^2}{4}\geq M_1^2$$ $$\sqrt[4]{\frac{3 M_2^2M_1^2+M_0^4}{4}}\geq M_1$$ I recognized and the terms for the arithmetic mean so the expression is now scale-invariant (independent on the nurmalization of the original set of numbers).

What we got is a successive averaging of different means. Let's call $f_k$ the function that computes the $k$-th mean of its arguments. First compute means $M_0=f_0(a,b,\ldots)$, $M_1$ and $M_2$ (analogously). Then apply additional averaging in the following order:

$$f_4(f_4(f_0(M_1,M_2),M_0),f_0(M_1,M_2))\geq M_1$$

Whatever you make from this, I'm interested to know.

orion
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