An inequality problem inspired by one of Cody's facebook posts:
Let \(k,a,b,c\) be reals. Prove that, if \(|k|\le 1\), then \[a^2+b^2+c^2\ge ab+bc+kca\]
and find equality case.
2 years, 10 months ago
Equality at \(a=b=c=0\) only.
Log in to reply
Also, equality at \(a=b=c\) when \(k=1\).
Oh, true! I was distracted by the presence of an \(a^2\) term so I assumed \(a=0\) :(
Directly show that
\[ a^2 + b^2 + c^2 \geq |ab | + |bc | + |ca| \geq ab+bc+kca. \]
In this case, the existence of \(k\) is a huge red herring.
Equality holds in the first with \( a = b =c \), and in the second with \( |ab| = ab, |bc| = bc, |ca| = kca \).
If \( k \neq \pm 1 \), then we must have \( ca = 0 \) which means \( a=b=c = 0 \).
If \( k = 1 \), we have \( a = b = c \) as the equality conditions.
If \( k = -1 \), we have \( ca \leq 0 \). But since \( a = b = c \), we must have \( a = b =c = 0 \).