Gheorghe Stoica's Problem from Gazeta Matematica


Gheorghe Stoica's Problem from Gazeta Matematica, problem

Solution 1

Let $M\,$ denote the required range. Let $a+b+c=3s\,$ and $ab+bc+ca=3(s^2-t^2),\,$ with $s,t\in\mathbb{R}.\,$ The given inequality becomes $s^2(k+1)+t^2(2-k)\ge 0.\,$ Observe that if $k\in [-1,2]\,$ then $f(s,t)=s^2(k+1)+t^2(2-k)\ge 0,\,$ implying $[-1,2]\subset M.$

On the other hand, if $k\lt -1\,$ then $f(1,0)\ge 0\,$ gives $k+1\ge 0\,$ - a contradiction.

If $k\ge 2\,$ then $f(0,1)\ge 0\,$ leads to $2-k\ge 0\,$ - a contradiction.

It follows that $M=[-1,2].$

Solution 2

Note that

$\displaystyle a^2+b^2+c^2+k(ab+bc+ca)=(\begin{array}\,a&b&c\end{array})\left(\begin{array}{aaa}\,1&\frac{k}{2}&\frac{k}{2}\\\frac{k}{2}&1&\frac{k}{2}\\\frac{k}{2}&\frac{k}{2}&1\end{array}\right)\left(\begin{array}{c}a\\b\\c\end{array}\right).$

The matrix in the above is diagonizable, with an eigenvalue $k+1\,$ and a double eigenvalue $\displaystyle 1-\frac{k}{2}.\,$ As a consequence, after a change of bases, $a^2+b^2+c^2+k(ab+bc+ca)\ge 0\,$ is equivalent to, say,

$\displaystyle (k+1)u^2+\left(1-\frac{k}{2}\right)v^2+\left(1-\frac{k}{2}\right)w^2\ge 0.$

Clearly, the above only holds for all $u,v,w\in\mathbb{R}\,$ if $k+1\ge 0\,$ and $\displaystyle 1-\frac{k}{2}\ge 0,\,$ i.e. when $k\in [-1,2].$

Solution 3

We have $\displaystyle k\ge -\frac{a^2+b^2+c^2}{ab+bc+ca}.\,$ Let $f(a,b,c)\,$ denote the right-hand side. The extrema of the function comes from the derivative equation $\nabla=0\,$ where

$\displaystyle \nabla=\left(\begin{array}{c}\,\displaystyle \frac{(a+b+c)(a(b+c)-b^2-c^2)}{(ab+bc+ca)^2}\\ \displaystyle \frac{(a+b+c)(b(c+a)-c^2-a^2)}{(ab+bc+ca)^2}\\ \displaystyle \frac{(a+b+c)(c(a+b)-a^2-b^2)}{(ab+bc+ca)^2} \end{array}\right).$

Solutions $a+b+c=0\,$ and $a=b=c,\,$ when substituted into the original inequality, give

$\displaystyle 2(a^2+ab+b^2)-k(a^2+ab+b^2)\ge 0,$

for $a+b+c=0\,$ and $3a^2+3ka^2\,$ for $a=b=c.\,$ From the former, $k\le 2\,$ and from the latter, $k\ge -1.$

Solution 4

For $k\in [-1,2],\,$ there is $\lambda\in [0,1]\,$ so that $k=\lambda\cdot(-1)+(1-\lambda)\cdot 2.\,$ We have:

$\displaystyle\begin{align} &a^2+b^2+c^2+k(ab+bc+ca)\\ &\qquad\qquad=\lambda(a^2+b^2+c^2)+(1-\lambda)(a^2+b^2+c^2)\\ &\qquad\qquad\qquad\qquad+[\lambda\cdot(-1)+(1-\lambda)\cdot 2](ab+bc+ca)\\ &\qquad\qquad=\lambda(a^2+b^2+c^2-ab-bc-ca)\\ &\qquad\qquad\qquad\qquad+(1-\lambda)[a^2+b^2+c^2+2(ab+bc+ca)]\\ &\qquad\qquad=\lambda\cdot\frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]\\ &\qquad\qquad\qquad\qquad+(1-\lambda)(a+b+c)^2\\ &\qquad\qquad\ge 0. \end{align}$

Solution 5

We'll verify that the matrix

$\displaystyle M=\left(\begin{array}{aaa}\,1&\frac{k}{2}&\frac{k}{2}\\\frac{k}{2}&1&\frac{k}{2}\\\frac{k}{2}&\frac{k}{2}&1\end{array}\right)$

is positive definite using Sylvester's theorem. $1\ge 0.\,$

$\displaystyle\left|\begin{array}{aa}\,1&\frac{k}{2}\\\frac{k}{2}&1\end{array}\right|=1-\left(\frac{k}{2}\right)^2\ge 0$

is equivalent to $k\in [-2,2].$

$\displaystyle\left|\begin{array}{aaa}\,1&\frac{k}{2}&\frac{k}{2}\\\frac{k}{2}&1&\frac{k}{2}\\\frac{k}{2}&\frac{k}{2}&1\end{array}\right|=(1+k)(2-k)^2\ge 0$

is equivalent to $k\ge -1.\,$ So that the matrix is positive definite for $k\in [-1,2].$

Solution 6

Letting $a=b=c=1\,$ we obtain $k\ge -1.\,$ Letting $a=1,b=-1,c=0\,$ we obtain $k\le 2.\,$

The function $f(k)=k(ab+bc+ca)+(a^2+b^2+c^2)\,$ is linear, also $f(-1)\ge 0\,$ and $f(2)\ge 0,\,$ implying $f(k)\ge 0\,$ for $k\in [-1,2].$

Solution 7

We look at the LHS of the required inequality as a function of $a:


For $f(a)\ge 0\,$ to hold for all $b\,$ and $c,\,$ it is necessary to have a non-negative discriminant: $k^2(b+c)^2-4(b^2+kbc+c^2)\le 0.\,$ Rewrite this as a function of $b:$


Observe that, in order to have $F(b)\le 0,\,$ we need $k^2-4\le 0\,$ and also

$\begin{align} 0\;&\ge 4c^2[k^2(k-2)^2-(k^2-4)^2]\\ &=-16c^2(k-2)^2(k+1). \end{align}$

The first one holds for $-2\le k\le 2;\,$ the second for $k\ge -1.$ The combination yields $-1\le k\le 2.$


Leo Giugiuc has kindly posted the above problem at the CutTheKnotMath facebook page messaged his solution (Solution 1), along with a link to a solution by Dimitris Kastriotis (Solution 2). To verify the latter I used an online matrix calculator WIMS. Solution 3 is by N. N. Taleb; Solutions 4 and 5 are by Marian Dinca; Solution 6 is by George Duca; Solution 7 is by Kunihiko Chikaya.


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