Good Inequality Problem

Given real numbers $a, b$ and $c$ so that $a^2 + b^2 + c^2 = k > 0.$ Then prove following inequality:

$\frac{2}{k}abc - \sqrt{2k} \leq a + b + c \leq \frac{2}{k}abc + \sqrt{2k}$

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

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}`	$\boxed{123}$

Comments

Here's my solution:

Without loss of generality, assume that $|b|\leq |c|\leq |a|$ . Then by $Cauchy - Schwarz$ Inequaility we have

$\begin{aligned} \left(a+b+c-\frac{2}{k}abc\right)^2 &= \left[a\left(1-\frac{2}{k}bc\right) + (b+c)\right]^2 \\ &\leq [a^2+(b+c)^2]\left[\left(1-\frac{2}{k}bc\right)^2 +1^2\right]\\ &= (k + 2bc) \left(2 - \frac{4}{k}bc +\frac{4}{k^2}b^2c^2\right) \\ &= 2k +\frac{8}{k^2} b^3c^3 - \frac{4}{k}b^2c^2\\ &= 2k +\frac{4}{k^2}b^2c^2(2bc - k) \end{aligned}$ Since $|b|\leq |c|\leq |a|$ , $a^2 \geq\frac{k}{3}$ , which implies that $\begin{aligned} 2bc \leq b^2 + c^2 = k - a^2 \leq k - \frac{k}{3} \leq \frac{2k}{3}< k \end{aligned}$ Therefore, $2bc - k < 0$ , which yields $\left(a+b+c-\frac{2}{k}abc\right)^2 \leq 2k$ or $\begin{aligned} \frac{2}{k}abc - \sqrt{2k}\leq a+b+c\leq \frac{2}{k}abc +\sqrt{2k} \end{aligned}$

Russelle Guadalupe - 7 years, 10 months ago

Great solution. Just to state that $2bc \leq b^2+c^2 < k$ since $a$ cannot be zero otherwise all of them are 0, which contradicts the condition $k>0$ , just showing this so we do not need the $\frac{k}{3}$ thing.

Yong See Foo - 7 years, 10 months ago

Good solution. Your main observation is that the inequality is equivalent to showing that $\left[ (a+b+c)(a^2+b^2+c^2) - 2abc \right]^2 \leq 2 (a^2+b^2 + c^2 )^3.$

This could also be done through brute force expansion and then AM-GM several times (i.e. Muirhead).

Calvin Lin Staff - 7 years, 10 months ago

Hands down, your solution answered it all. Thanks :)

Fariz Azmi Pratama - 7 years, 10 months ago

Hi, if we substitute $k = a^2 + b^2 + c^2$ into the inequalities, then the system of inequalities is homogeneous, let $u = a + b + c$ , $v = ab + bc + ca$ and $w = abc$ , then the system becomes $\frac{2w}{\sqrt{u^2 - 2v}} - \sqrt{2u^2 - 4v} \leq u \leq \frac{2w}{u^2 - 2v} + \sqrt{2u^2 - 4v}$ i.e. $\frac{2w}{\sqrt{u^2 - 2v}} - \sqrt{2u^2 - 4v} - u \leq 0 \leq \frac{2w}{u^2 - 2v} + \sqrt{2u^2 - 4v} - u$ Both expressions are linear in $w$ . By $uvw$ method it suffices to prove inequality in two cases 1) $(a - b)(b - c)(c - a) = 0$ , 2) $abc = 0$ , in the first case we let WLOG $b = c$ and we have to prove $\frac{2ab^2}{a^2 + 2b^2} - \sqrt{2a^2 + 4b^2} \leq a + 2b \leq \frac{2ab^2}{a^2 + 2b^2} + \sqrt{2a^2 + 4b^2}$ which is easy. In the second case WLOG let $c = 0$ and we have to prove $-\sqrt{2a^2 + 2b^2} \leq a + b \leq \sqrt{2a^2 + 2b^2}$ which is again easy.

Here is link to the method used: PDF I used different substitution, but it doesn't interfere with the method.

Jan J. - 7 years, 10 months ago

It took me a bit longer to understand, but I got it. Thanks :)

Fariz Azmi Pratama - 7 years, 10 months ago

Clearly the leftmost side is less than the rightmost side as $k$ is positive. But beyond that, I'll have to do more than a first time glance to get this!

Tanishq Aggarwal - 7 years, 11 months ago

Yes, and to be honest I don't have any clue to solve this inequality.

Fariz Azmi Pratama - 7 years, 11 months ago

Any chance you could tell us where you got this problem?

Sotiri Komissopoulos - 7 years, 11 months ago

Oh sorry, my friend gave me this problem. Unfortunately she didn't mention where she get this problem from.

Fariz Azmi Pratama - 7 years, 11 months ago

That's fine, thanks. :)

Sotiri Komissopoulos - 7 years, 10 months ago

for 3 real numbers a,b and c, notice that AM=(a+b+c)/3, GM=(abc)^(1/3) and RMS=sqrt[(a^2+b^2+c^2)/3]. try to use RMS>AM>GM.i tried and it worked.but the solution is around 20-22 lines long.so will take a long time to type :-(

Adeeb Zaman - 7 years, 10 months ago

How do you account for the fact that a,b, and/or c could be negative numbers too. I thought AM-GM only works for positive numbers.

Edwin Ma - 7 years, 10 months ago

Wow that is so long. You did it great. But I think for this problem, $AM - GM$ has nothing to do here. Like what David say, $AM - GM$ only works on positive number. Let say for your case $a = 1, b = -1$ and $c = 0$ then $AM = 0 < GM = \sqrt[3]{2}.$

Fariz Azmi Pratama - 7 years, 10 months ago

Put the value of $k$ in both sides, square up, rearrange & you will get it after some reasoning...

A Brilliant Member - 7 years, 11 months ago

What do you mean by put the value of $k$ in both sides? Since there are three sides on the ineq.

Fariz Azmi Pratama - 7 years, 10 months ago

easy

Youssef Ihizem - 7 years, 11 months ago

Would you like to show your solution or maybe some hints?

Fariz Azmi Pratama - 7 years, 10 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}`	$\boxed{123}$