Cool inequality #0 Making it easy

Algebra Level 5

If $a, b,$ and $c$ are real numbers such that $a^2 + b^2 + c^2 = 1$ , then the value of $ab+bc+ca$ is on the interval $[A, B]$ where $A$ is as large as possible and $B$ is as small as possible.

Find $|A| + |B|$ .

The answer is 1.5.

1 solution

Department 8
Jul 13, 2015

Case I $a+b+c\ge 0$

For this

${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2(ab+bc+ca)\ge 0\\1+2(ab+bc+ca)\ge 0 \\ab+bc+ca\ge -0.5$

Case 2 ${ (a-b) }^{ 2 }\ge 0\\ { (b-c) }^{ 2 }\ge 0\\ { (c-a) }^{ 2 }\ge 0$

First Squaring and adding all $2({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca)\ge 0\\ 1-ab-bc-ca\ge 0\\ 1\ge ab+bc+ca$

This means

$-0.5\le ab+bc+ca\le 1$

Answer = $|-0.5|+|1|$ = $\boxed{1.5}$

Note :- The method may be wrong but the answer is correct

There are 3 exactly same question i have saw on brilliant

Aakash Khandelwal - 5 years, 7 months ago

Me too one level 3, level 4, level 5 (these are what I saw)

Department 8 - 5 years, 7 months ago

Cool inequality #0 Making it easy

The answer is 1.5.

1 solution

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