Good Inequality Problem

Given real numbers a,ba, b and cc so that a2+b2+c2=k>0.a^2 + b^2 + c^2 = k > 0. Then prove following inequality:

2kabc2ka+b+c2kabc+2k\frac{2}{k}abc - \sqrt{2k} \leq a + b + c \leq \frac{2}{k}abc + \sqrt{2k}

Note by Fariz Azmi Pratama
5 years, 11 months ago

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Here's my solution:

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

(a+b+c2kabc)2=[a(12kbc)+(b+c)]2[a2+(b+c)2][(12kbc)2+12]=(k+2bc)(24kbc+4k2b2c2)=2k+8k2b3c34kb2c2=2k+4k2b2c2(2bck)\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 bca|b|\leq |c|\leq |a| , a2k3a^2 \geq\frac{k}{3}, which implies that 2bcb2+c2=ka2kk32k3<k\begin{aligned} 2bc \leq b^2 + c^2 = k - a^2 \leq k - \frac{k}{3} \leq \frac{2k}{3}< k \end{aligned} Therefore, 2bck<02bc - k < 0, which yields (a+b+c2kabc)22k\left(a+b+c-\frac{2}{k}abc\right)^2 \leq 2k or 2kabc2ka+b+c2kabc+2k\begin{aligned} \frac{2}{k}abc - \sqrt{2k}\leq a+b+c\leq \frac{2}{k}abc +\sqrt{2k} \end{aligned}

Russelle Guadalupe - 5 years, 11 months ago

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Great solution. Just to state that 2bcb2+c2<k2bc \leq b^2+c^2 < k since aa cannot be zero otherwise all of them are 0, which contradicts the condition k>0k>0, just showing this so we do not need the k3\frac{k}{3} thing.

Yong See Foo - 5 years, 11 months ago

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Good solution. Your main observation is that the inequality is equivalent to showing that [(a+b+c)(a2+b2+c2)2abc]22(a2+b2+c2)3. \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 - 5 years, 10 months ago

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

Fariz Azmi Pratama - 5 years, 11 months ago

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Hi, if we substitute k=a2+b2+c2k = a^2 + b^2 + c^2 into the inequalities, then the system of inequalities is homogeneous, let u=a+b+cu = a + b + c, v=ab+bc+cav = ab + bc + ca and w=abcw = abc, then the system becomes 2wu22v2u24vu2wu22v+2u24v\frac{2w}{\sqrt{u^2 - 2v}} - \sqrt{2u^2 - 4v} \leq u \leq \frac{2w}{u^2 - 2v} + \sqrt{2u^2 - 4v} i.e. 2wu22v2u24vu02wu22v+2u24vu\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 ww. By uvwuvw method it suffices to prove inequality in two cases 1) (ab)(bc)(ca)=0(a - b)(b - c)(c - a) = 0, 2) abc=0abc = 0, in the first case we let WLOG b=cb = c and we have to prove 2ab2a2+2b22a2+4b2a+2b2ab2a2+2b2+2a2+4b2\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=0c = 0 and we have to prove 2a2+2b2a+b2a2+2b2-\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. - 5 years, 11 months ago

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

Fariz Azmi Pratama - 5 years, 11 months ago

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

Tanishq Aggarwal - 5 years, 11 months ago

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

Fariz Azmi Pratama - 5 years, 11 months ago

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

Sotiri Komissopoulos - 5 years, 11 months ago

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

Fariz Azmi Pratama - 5 years, 11 months ago

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That's fine, thanks. :)

Sotiri Komissopoulos - 5 years, 11 months ago

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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 :-(

Adeeb Zaman - 5 years, 11 months ago

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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.

Edwin Ma - 5 years, 11 months ago

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Wow that is so long. You did it great. But I think for this problem, AMGMAM - GM has nothing to do here. Like what David say, AMGMAM - GM only works on positive number. Let say for your case a=1,b=1a = 1, b = -1 and c=0c = 0 then AM=0<GM=23.AM = 0 < GM = \sqrt[3]{2}.

Fariz Azmi Pratama - 5 years, 11 months ago

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

A Brilliant Member - 5 years, 11 months ago

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

Fariz Azmi Pratama - 5 years, 11 months ago

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easy

Youssef Ihizem - 5 years, 11 months ago

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

Fariz Azmi Pratama - 5 years, 11 months ago

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