Inspired by Cody

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.

Note by Daniel Liu
3 years, 9 months ago

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Ah,

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

Equality at $$a=b=c=0$$ only.

- 3 years, 9 months ago

Also, equality at $$a=b=c$$ when $$k=1$$.

- 3 years, 9 months ago

Oh, true! I was distracted by the presence of an $$a^2$$ term so I assumed $$a=0$$ :(

- 3 years, 9 months ago

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 $$a = b =c = 0$$.

Staff - 3 years, 9 months ago