# Proving a weird inequality

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. :)

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

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

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

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