3
$\begingroup$

Recently I saw this interesting inequality problem from Singapore Math Olympiad 2016 (Open Section, Special Round), but I am not sure how to start, here it goes:

Real numbers $a,b,c$ are such that $0<a,b,c<\frac{1}{2}$ and $a+b+c=1.$ Prove that for all real numbers $x,y,z,$ we have $$abc(x+y+z)^2\geq a(1-2a)yz+b(1-2b)xz+c(1-2c)xy.$$

I have tried to use Rearrangement inequality but it doesn't really work, as it boils down to $3bc\geq 1-2a$ and this statement fails when $a=b=0.3, c=0.4.$ Any thoughts?


P.S. I would prefer hints instead of full solutions because this will serve as one of my preparatory exercises before I go for the exact same competition mentioned above later in June this year. Also, this is my first time using this platform, happy to learn from you guys. :)

$\endgroup$
3
$\begingroup$

$a+b-c=1-2c>0$, $a+c-b>0$ and $b+c-a>0$,

which says that $a$, $b$ and $c$ they are sides-lengths of a triangle and we need to prove that $$abc(x+y+z)^2\geq(a+b+c)(a(b+c-a)yz+b(a+c-b)xz+c(a+b-c)xy)$$ or $$(x+y+z)^2\geq\frac{(b+c)^2-a^2}{bc}\cdot yz+\frac{(a+c)^2-b^2}{ac}\cdot xz+\frac{(a+b)^2-c^2}{ab}\cdot xy$$ or $$x^2+y^2+z^2\geq2yz\cos\alpha+2xz\cos\beta+2xy\cos\gamma.$$ Now, let $Z\in AB$, $X\in BC$ and $Y\in CA$, such that $\vec{AZ}=\frac{z}{c}\vec{AB},$ $\vec{BX}=\frac{x}{a}\vec{BC}$ and $\vec{CY}=\frac{y}{b}\vec{CA}.$ Thus, the last inequality it's just $$\left(\vec{AZ}+\vec{BX}+\vec{CY}\right)^2\geq0.$$ Done!

$\endgroup$
  • $\begingroup$ I wonder if this qualifies as "hints instead of full solutions" ... $\endgroup$ – Martin R Jun 18 '18 at 8:47
0
$\begingroup$

Analyzing the inequality, it's cyclic aspect suggest the utilization of

$$ a = b = c = \frac 13 $$

Substituting we obtain

$$ f(x,y,z) = x^2+y^2+z^2 -z x-z y - x y $$

and the associated hessian is

$$ H = \left( \begin{array}{ccc} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \\ \end{array} \right) $$

with eigenvalues $(3,3,0)$ so it guarantees

$$ f(x,y,z) \ge 0 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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