An Inequality with Constraint XVII

Statement

an inequality with constraints, #17

Solution 1

We'll solve an equivalent problem:

For $a,b,c\ge 0,\,$ such that $a+b+c=\sqrt[3]{4}+2,\,$ we have $(a-1)^2+(b-1)^2+(c-1)^2\ge 0.$

Let's denote $\sqrt[3]{4}+2=3s\gt 3,\,$ $ab+bc+ca=3(s^2-t^2),\,$ where $0\le t\le s\,$ and $abc=p.\,$ We get $a^3+b^3+c^3=(9t^2s+p),\,$ $a^2+b^2+c^2=3(s^2+2t^2).\,$ Hence, we need to prove that

(*)

$9t^2s+p-3(s^2+2t^2)-3s+1\ge 0.$

We know that

$\displaystyle\min (p)=\begin{cases} -2t^3-3t^2s+s^3,&if\;0\le t\frac{s}{2}\\ 0,&if\,\frac{s}{2}\le t\le s. \end{cases}$

Suffice it to prove (*) for $p\,$ being the minimum, $\min (p).\,$ If $\displaystyle 0\le t\frac{s}{2},\,$ we have to show that

$\displaystyle s^3+6t^2s-2t^3-3s^2-6t^2+3s-1\ge 0.$

However, since $2\gt s\gt 1,\,$ the function $f(t)=6t^2s-2t^3-6t^2,\,$ has a maximum on $\displaystyle \left(0,\frac{s}{2}\right),\,$ implying $\min f=\displaystyle f\left(\frac{s}{2}\right),\,$ hence, setting, for the typographical simplicity sake, $k=\sqrt[3]{4}+2,$

$\displaystyle \begin{align} s^2+6t^2s-2t^3-3s^2&-6t^2+3s-1\ge\frac{1}{4}(9s^3-18s^2+12s-4)\\ &=\frac{1}{12}[(k-2)^3-4]=0. \end{align}$

The case $\displaystyle \frac{s}{2}\le t\le s\,$ is trivial.

Solution 2

Set $k=\sqrt[3]{4}-1.\,$ Note that $k\gt 0.\,$ Assume, WLOG, $a\ge b\ge c\ge -1.\,$ We shall consider several cases.

Case 1: $0\ge a\ge b\ge c\ge -1.\,$

$0\ge a+b+c=k\gt 0.\,$ Contradiction.

Case 2: $a\ge b\ge c\ge 0.\,$

There's nothing to prove: $a^3+b^3+c^3\ge 0.$

Case 3: $a\ge 0\ge b\ge c\ge -1.\,$

From $a+b+c=k\gt 0,\,$ $a\gt -(b+c),\,$ such that $a^3\gt -(b+c)^3,\,$ implying $a^3+b^3+c^3\gt -3bc(b+c)\ge 0,\,$ because $b\le 0\,$ and $c\le 0.$

Case 4: $a\ge b\ge 0\ge c\ge -1.\,$

Since $0\ge c=k-a-b\ge -1,\,$ $k+1\ge a+b\ge k.\,$ Set $a+b=s,\,$ $ab=p.\,$ We have

$0\le (a-b)^2=(a+b)^2-4ab=s^2-4p\,$

implying

(1)

$\displaystyle \frac{s^2}{4}\ge p$

and $k\le s\le k+1.\,$ Further,

$\begin{align} a^3+b^3+c^3 &= a^3+b^3+(k-a-b)^3\\ &=k^3-3k^2(a+b)+3k(a^2+b^2)-3ab(a+b)+6kab\\ &=k^3-3k^2s+3k(s^2-2p)-3sp+6kp\\ &=k^3-3k^2s+3ks^2-3sp. \end{align}$

It follows that, $a^3+b^3+c^3\ge 0\,$ is equivalent to

$\displaystyle \frac{k^3-3k^2s+3ks^2}{3s}\ge p.$

Due to (1), suffice it to prove

(2)

$\displaystyle \frac{k^3-3k^2s+3ks^2}{3s}\ge \frac{s^2}{4}.$

The latter is equivalent to

$T=3s^3-12ks^2+12k^2s-4k^3\le 0,$

which is true because $T=3(s-k-1)(s^2-s(3k-1)+(k-1)^2)\,$ while $s\le k+1\,$ and

$\Delta=(3k-1)^2-4(k-1)^2=(k+1)(5k-3)\lt 0$

because $\displaystyle 0\lt k\lt\frac{3}{5}.\,$ Thus (2) holds and this completes the proof.

Solution 3

Let $x=a+1,\,$ $y=b+1,\,$ $z=c+1.\,$ $x,y,z\ge 0,\;$ $r=xyz\ge 0,\,$ $p=x+y+z=2+2^{\frac{2}{3}}.\,$ Thus

$\displaystyle\begin{align} \sum_{cycl}a^3&=\sum_{cycl}(x-1)^3=\sum_{cycl}x^3-3\sum_{cycl}x^2+3\sum_{cycl}x-3\\ &=q^2(p-2)-p^2+3p-3+3r, \end{align}$

where $0\le q\le p\,$ and, according to Vo Quoc Ba Can's inequality, $\displaystyle r\ge \frac{1}{27}(p+q)^2(p-2q).$

Let $\displaystyle q_0=\frac{p}{2}\approx 1.7937.\,$ We shall consider two cases

Case 1: $\displaystyle q\gt\frac{p}{2}=q_0$

In this case, with $p=2+2^{\frac{2}{3}},$

$\displaystyle q^2(p-2)-p^2+3p-3\ge \frac{p^2}{4}(p-2)+3p-3=0,$

implying $a^3+b^3+c^3\gt 0.$ So done.

Case 2: $\displaystyle q\le q_0 = \frac{p}{2}$

$\displaystyle \sum_{cycl}a^3 \ge pq^2-p^2-2q^2+3p-3+\frac{1}{9}(p+q)^2(p-2q)=\frac{1}{9}f(q),$

where $f(q)=-2q^3-18q^2+6pq^2+(p-3)^3\,$ and $f'(q)=6q(-q-6+2p),\,$ with roots $0\,$ and $2p-6\lt p.\,$ Thus, for $0\le q\le 2p-6,\,$ $f'(q)\ge 0\,$ and $f'(q)\lt 0\,$ when $q\gt 2p-6.\,$ Thus the function is increasing to the left of $2p-6\,$ and decreasing afterwards. $f(0)=(p-3)^3\gt 0\,$ while also $\displaystyle f\left(\frac{p}{2}\right)\gt 0,\,$ so that $f(q)\gt 0\,$ for $q\in \left[0,\displaystyle \frac{p}{2}\right]\,$ and, to combine the two cases, it is positive for $q\in [0,p].$

Equality occurs for $q=q_0=\displaystyle \frac{p}{2}=1+2^{\frac{1}{3}}.\,$ When this happens, $r=0\,$ and $\displaystyle \sum_{cycl}xy=1+2^{-\frac{1}{3}}+2^{\frac{1}{3}}.\,$ $(x,y,z=(1+2^{-\frac{1}{3}},1+2^{-\frac{1}{3}})\,$ and permutations, i.e., for $(a,b,c)=(2^{-\frac{1}{3}},2^{-\frac{1}{3}},0)\,$ and permutations.

Solution 4

an inequality with constraints, #17, illustration

Acknowledgment

The problem and Solution 1 have been kindly communicated by Leo Giugiuc; Solution 2 is by Marian Cucoanes; Solution 3 is by Imad Zak; Solution 4 is by N. N. Taleb. The author of the problem is unknown.

 

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