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.

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.

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## Comments

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\[(a-b)^2+(b-c)^2+(c-ka)^2+(1-k^2)a^2\ge0\]

Equality at \(a=b=c=0\) only. – Cody Johnson · 2 years, 1 month ago

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– Daniel Liu · 2 years, 1 month ago

Also, equality at \(a=b=c\) when \(k=1\).Log in to reply

– Cody Johnson · 2 years, 1 month ago

Oh, true! I was distracted by the presence of an \(a^2\) term so I assumed \(a=0\) :(Log in to reply

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 \( a = b =c = 0 \). – Calvin Lin Staff · 2 years, 1 month ago

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