# Prove that $f(A)\leq max(f(P),f(Q),f(R))$

Consider any $\bigtriangleup PQR$ in the $x-y$ plane. Let $f(x,y)=ax+by+c$ , where $a,b,c\in\mathbb{R}$. Let $A\in\mathbb{R^2}$ be any point in the interior or on the $\bigtriangleup PQR$. Prove that $f(A)\leq max(f(P),f(Q),f(R))$ where $A,P,Q,R\in \mathbb{R^2}$.
Source- Indian Statistical Institute Entrance Exam, 2014
N.B.- Intuitively it seems correct but I don't know how to give a rigorous argument. Also please check if the "tag" and "title" of the question are appropriate.Thanks

It is easy to see that $f(x,y)=c_0$ for a constant $c_0$ defines a line. Particularly, $f(x,y)=f(A)$ defines a line. This line divides the plane in two parts: $f(x,y)<f(A)$ an $f(x,y)\geq f(A)$. It is easy to see, that the part $f(x,y)\geq f(A)$ contains at least one vertex of the $\bigtriangleup PQR$. QED.
• This is all good. Rather this was also in my mind. I wanted to know how do we prove that a line divides a plane into 2 parts such that $f(x,y)<f(A)$ in one part and $f(x,y)\geq f(A)$ in other. I mean how to prove that on one side of the line $f(x,y)>0$ and keeps on increasing in magnitude as we move away from the line? – idpd15 Jun 15 '14 at 13:37
• Firstly we should that points with different "signs" belong to different half-planes. Due to Intermediate value theorem for two points $(x_1,y_1)$ and $(x_2,y_2)$ with $f(x_1,y_1)>c_0$ and $f(x_2,y_2)<c_0$ there is a point $(x,y)$ on line segment $[(x_1,y_1),(x_2,y_2)]$ with $f(x_1,y_1)=c_0$, i.e, $(x,y)$ is on the line and hence $(x_1,y_1)$ and $(x_2,y_2)$ are in different half-planes. – IBazhov Jun 15 '14 at 18:05
• Secondly, we show that there are a half-plane with "<" and a half-plane with ">". Check that function $f(x,y)$ increases in direction $(a,b)$ and decreases in direction $-(a,b)$. Now if for a point $(x,y)$ we have $f(x,y)=c_0$ then for $(x+a,y+b)$ we have $f(x+a,y+b)>c_0$ and for $(x-a,y-b)$ we have $f(x-a,y-b)<c_0$. – IBazhov Jun 15 '14 at 18:32