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.

#Algebra #Factorization #Inequality #Proof #TrivialInequality

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

Ah,

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

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

Cody Johnson - 6 years, 6 months ago

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

Daniel Liu - 6 years, 6 months ago

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

Cody Johnson - 6 years, 6 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 $a = b =c = 0$ .

Calvin Lin Staff - 6 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$