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

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

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## Comments

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TopNewestHere'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 $\begin{aligned} \frac{2}{k}abc - \sqrt{2k}\leq a+b+c\leq \frac{2}{k}abc +\sqrt{2k} \end{aligned}$

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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.

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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).

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

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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.

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

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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!

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

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

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

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That's fine, thanks. :)

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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 :-(

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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.

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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 $AM = 0 < GM = \sqrt[3]{2}.$

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

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

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easy

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

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