Triangle Inequality is a starting point but that's a bit trivial

a2b2+2c2a2+b2c2+2a2b2+c2a2+2b2c23\dfrac {a}{\sqrt{2b^2+2c^2-a^2}} + \dfrac {b}{\sqrt{2c^2+2a^2-b^2}} + \dfrac {c}{\sqrt{2a^2+2b^2-c^2}} \geq \sqrt{3}

Let aa, bb and cc be the lengths of the sides of a triangle. Prove the inequality above.

Note by Sharky Kesa
3 years, 8 months ago

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Solution.\textbf{Solution.} Suffices to prove a2b2+2c2a23a2a2+b2+c2\frac{a}{\sqrt{2b^2+2c^2-a^2}} \geq \frac{\sqrt{3}a^2}{a^2 + b^2 + c^2} which is just (2a2+b2+c2)20(-2a^2 + b^2 + c^2)^2 \geq 0.

We finish by adding cyclically. \square

Alan Yan - 3 years, 8 months ago

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I am not able to prove it can you please help

Department 8 - 3 years, 8 months ago

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Just expand and simplify the expression I wrote.

Alan Yan - 3 years, 8 months ago

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Can u prove that median's relation I assumed from the proof statement and written in reply to Calvin Lin?

Shyambhu Mukherjee - 3 years, 8 months ago

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How would one think of coming up with that first equation?

Calvin Lin Staff - 3 years, 8 months ago

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This concept, called fudging\textbf{fudging}, is a well-known technique in inequality solving. However, in my opinion, it is the hardest technique to implement. It can usually only be achieved by doing a lot of inequality problems. The whole point is to get each term under the cyclic summation to have a common denominator in order to finish simultaneously. Yufei Zhao's document addresses a calculus approach to find the right "fudging" expression if you can't see it immediately.

http://yufeizhao.com/wc08/ineq.pdf

Alan Yan - 3 years, 8 months ago

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cosA=b2+c2a22bc(cosine..rule)cosA=\frac{b^{2}+c^{2}-a^{2}}{2bc} (cosine..rule)
we know that cosθ1cos\theta\leq1. Therefore
1b2+c2a22bc1\geq\frac{b^{2}+c^{2}-a^{2}}{2bc}
2bc+b2+c22b2+2c2a22bc+b^{2}+c^{2}\geq2b^{2}+2c^{2}-a^{2} 2bc+b2+c22b2+2c2a2\sqrt {2bc+b^{2}+c^{2}}\geq\sqrt{2b^{2}+2c^{2}-a^{2}} ab+ca2b2+2c2a2\frac{a}{b+c}\leq\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}

Therefore we get cycab+ccyca2b2+2c2a2\sum_{cyc}\frac{a}{b+c}\leq\sum_{cyc}\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}} By using Nesbitts Inequality we get 32cycab+ccyca2b2+2c2a2\frac{3}{2}\leq\sum_{cyc}\frac{a}{b+c}\leq\sum_{cyc}\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}} Hence 32cyca2b2+2c2a2\frac{3}{2}\leq\sum_{cyc}\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}.

I am not getting the flaw I made in my solution please help me.

Shivam Jadhav - 3 years, 8 months ago

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Is the degenerate triangle excluded in the problem?

Reynan Henry - 3 years, 8 months ago

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Typically, the degenerate triangle is excluded from being a triangle.

In this case, I don't think it matters. The inequality is true for {a,b,c}={0,1,1} \{ a, b, c \} = \{ 0, 1, 1 \} .

Calvin Lin Staff - 3 years, 8 months ago

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@Calvin Lin Sir my account has been hacked can you block my account from all the places where I have logged in. Please we can talk on my email

Department 8 - 3 years, 8 months ago

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@Department 8 You can change your password which will log you out of all other devices.

Calvin Lin Staff - 3 years, 8 months ago

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There is no flaw in the solution. You have shown that the LHS is greater than 1.5, which is a correct statement. You just have not shown that the LHS is greater that 31.7 \sqrt{3} \approx 1.7 .

For example, if we wanted to prove that x2+12x x^2 + 1 \geq 2x , just because our steps showed that x2+12x1000 x^2 + 1 \geq 2x - 1000 , doesn't mean that our solution has a flaw. It just means that our solution doesn't lead to the desired conclusion as yet.

FYI, be aware of the spacing that you use, and check that it is displaying as you would like it to. I've edited your solution to make it more readable.

Calvin Lin Staff - 3 years, 8 months ago

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Can you suggest the method to solve this question or what I need to add to my solution to make it valid .

Shivam Jadhav - 3 years, 8 months ago

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Is there any inequality related to the length of a side and median upon it? Actually if we analytically say that length of median upon a side lengthed x is less than equal to √3x/2 then putting the median length in terms of sides we get each part of this inequality greater than equal to 1/√3. So our inequality gets solved.

Shyambhu Mukherjee - 3 years, 8 months ago

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@Shyambhu Mukherjee If I have not made any computational errors, the measure of the A-median would be ma=2b2+2c2a22m_a = \frac{\sqrt{2b^2 + 2c^2 - a^2}}{2} . Your proposal is to prove that 2b2+2c2a223a2    2b2+2c2a23a2    b2+c22a2\frac{\sqrt{2b^2 + 2c^2 - a^2}}{2} \leq \frac{\sqrt{3}a}{2} \implies 2b^2 + 2c^2 - a^2 \leq 3a^2 \implies b^2 + c^2 \leq 2a^2 which is obviously false given a triangle such that b,c>ab , c > a.

Alan Yan - 3 years, 8 months ago

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@Alan Yan Thanks . actually I guessed it only. Bad guess I guess.

Shyambhu Mukherjee - 3 years, 8 months ago

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I have added another solution just help me to finish it.

Shivam Jadhav - 3 years, 8 months ago

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I'm stuck. Is there anyone to correct this solution? Assume the sides are a2a^2, b2b^2, c2c^2 sides, then 2s=a2+b2+c22s = a^2 + b^2 +c^2 . Rearranging these terms, 14s2a23+14s2b23+14s2c23\sqrt{\frac{1}{4\frac{s^2}{a^2}-3}}+\sqrt{\frac{1}{4\frac{s^2}{b^2}-3}}+\sqrt{\frac{1}{4\frac{s^2}{c^2}-3}} Using AM-HM, 14sa23+14sb23+14sc2394sa23+4sb23+4sc23\sqrt{\frac{1}{4\frac{s}{a^2}-3}}+\sqrt{\frac{1}{4\frac{s}{b^2}-3}}+\sqrt{\frac{1}{4\frac{s}{c^2}-3}} \geq \frac{9}{\sqrt{4\frac{s}{a^2}-3}+\sqrt{4\frac{s}{b^2}-3}+\sqrt{4\frac{s}{c^2}-3}} Consider the denumerator and using Cauchy-Schwarz, 4sa23+4sb23+4sc233(4s(1a2+1b2+1c2)9)\sqrt{4\frac{s}{a^2}-3}+\sqrt{4\frac{s}{b^2}-3}+\sqrt{4\frac{s}{c^2}-3} \leq \sqrt{3(4s(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})-9)} It affects that 94sa23+4sb23+4sc2393(4s(1a2+1b2+1c2)9)\frac{9}{\sqrt{4\frac{s}{a^2}-3}+\sqrt{4\frac{s}{b^2}-3}+\sqrt{4\frac{s}{c^2}-3}}\geq \frac{9}{\sqrt{3(4s(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})-9)}}

I'm stuck. Is there any flaw in my solution?

Figel Ilham - 3 years, 8 months ago

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For those who want a hint, try an incenter substitution.

That is, let a=x+ya = x+y, b=y+zb=y+z and c=z+xc=z+x. Note that xx, yy and zz are positive reals.

Sharky Kesa - 3 years, 8 months ago

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Ummm... any hint?

Department 8 - 3 years, 8 months ago

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Using Titus..lemmaTitu's..lemma we get cyca2b2+2c2a2(a+b+c)22b2+2c2a2\sum_{cyc}\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\geq\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}{\sqrt{2b^{2}+2c^{2}-a^{2}}} By using RMSAMRMS-AM inequality we get 2b2+2c2a2+2c2+2a2b2+2b2+2a2c232b2+2c2a23\sqrt{\frac{2b^{2}+2c^{2}-a^{2}+2c^{2}+2a^{2}-b^{2}+2b^{2}+2a^{2}-c^{2}}{3}}\geq\frac{\sqrt{2b^{2}+2c^{2}-a^{2}}}{3} This implies 3a2+b2+c22b2+2c2a23\sqrt{a^{2}+b^{2}+c^{2}}\geq \sqrt{2b^{2}+2c^{2}-a^{2}} (a+b+c)22b2+2c2a2(a+b+c)23a2+b2+c2(ab+bc+ca)a2+b2+c2\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\geq\frac{ (\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}{3\sqrt{a^{2}+b^{2}+c^{2}}}\geq\frac{ (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{\sqrt{a^{2}+b^{2}+c^{2}}} Therefore cyca2b2+2c2a2(ab+bc+ca)a2+b2+c2\sum_{cyc}\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\geq\frac{(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{\sqrt{a^{2}+b^{2}+c^{2}}} Now anyone please help me to finish the solution.

Shivam Jadhav - 3 years, 8 months ago

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Unfortunately, the RHS of your last equation is not 3 \geq \sqrt{3} . For example, substitute in a=0+,b=1,c=1 a = 0^+, b = 1, c = 1 , and we obtain 12 \approx \frac{1}{\sqrt{2} } .

In using 3a2+b2+c22b2+2c2a2 3 \sqrt{ a^2 + b^2 + c^2 } \geq \sqrt{ 2 b^2 + 2c^2 - a ^2 } you made it way too loose, which is why the bound does not work. For example, we have2a2+b2+c22a2+2b2+2c2>2b2+2c2a2 \sqrt{2} \sqrt{ a^2 + b^2 + c^2 } \geq \sqrt{2 a^2 + 2b^2 + 2c^2 } > \sqrt{ 2 b^2 + 2c^2 - a^2} .

Calvin Lin Staff - 3 years, 8 months ago

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Sir, please can you explain me the "bound" thing??

Pranjal Mittal - 3 years, 8 months ago

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