\[\dfrac {a}{\sqrt{2b^2+2c^2-a^2}} + \dfrac {b}{\sqrt{2c^2+2a^2-b^2}} + \dfrac {c}{\sqrt{2a^2+2b^2-c^2}} \geq \sqrt{3}\]

Let \(a\), \(b\) and \(c\) be the lengths of the sides of a triangle. Prove the inequality above.

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## Comments

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TopNewest\(\textbf{Solution.}\) Suffices to prove \[\frac{a}{\sqrt{2b^2+2c^2-a^2}} \geq \frac{\sqrt{3}a^2}{a^2 + b^2 + c^2} \] which is just \((-2a^2 + b^2 + c^2)^2 \geq 0\).

We finish by adding cyclically. \(\square\)

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I am not able to prove it can you please help

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Just expand and simplify the expression I wrote.

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

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

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This concept, called \(\textbf{fudging}\), 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

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\[cosA=\frac{b^{2}+c^{2}-a^{2}}{2bc} (cosine..rule)\]

we know that \(cos\theta\leq1\). Therefore

\[1\geq\frac{b^{2}+c^{2}-a^{2}}{2bc}\]

\[2bc+b^{2}+c^{2}\geq2b^{2}+2c^{2}-a^{2}\] \[\sqrt {2bc+b^{2}+c^{2}}\geq\sqrt{2b^{2}+2c^{2}-a^{2}}\] \[\frac{a}{b+c}\leq\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\]

Therefore we get \[\sum_{cyc}\frac{a}{b+c}\leq\sum_{cyc}\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\] By using Nesbitts Inequality we get \[\frac{3}{2}\leq\sum_{cyc}\frac{a}{b+c}\leq\sum_{cyc}\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\] Hence \[\frac{3}{2}\leq\sum_{cyc}\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\].

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

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Is the degenerate triangle excluded in the problem?

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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 \( \{ a, b, c \} = \{ 0, 1, 1 \} \).

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change your password which will log you out of all other devices.

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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 \( \sqrt{3} \approx 1.7 \).

For example, if we wanted to prove that \( x^2 + 1 \geq 2x \), just because our steps showed that \( x^2 + 1 \geq 2x - 1000 \), 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.

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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 .

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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.

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

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I'm stuck. Is there anyone to correct this solution? Assume the sides are \(a^2\), \(b^2\), \(c^2\) sides, then \(2s = a^2 + b^2 +c^2 \). Rearranging these terms, \[\sqrt{\frac{1}{4\frac{s^2}{a^2}-3}}+\sqrt{\frac{1}{4\frac{s^2}{b^2}-3}}+\sqrt{\frac{1}{4\frac{s^2}{c^2}-3}}\] Using AM-HM, \[\sqrt{\frac{1}{4\frac{s}{a^2}-3}}+\sqrt{\frac{1}{4\frac{s}{b^2}-3}}+\sqrt{\frac{1}{4\frac{s}{c^2}-3}} \geq \frac{9}{\sqrt{4\frac{s}{a^2}-3}+\sqrt{4\frac{s}{b^2}-3}+\sqrt{4\frac{s}{c^2}-3}}\] Consider the denumerator and using Cauchy-Schwarz, \[\sqrt{4\frac{s}{a^2}-3}+\sqrt{4\frac{s}{b^2}-3}+\sqrt{4\frac{s}{c^2}-3} \leq \sqrt{3(4s(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})-9)}\] It affects that \[\frac{9}{\sqrt{4\frac{s}{a^2}-3}+\sqrt{4\frac{s}{b^2}-3}+\sqrt{4\frac{s}{c^2}-3}}\geq \frac{9}{\sqrt{3(4s(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})-9)}}\]

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

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

That is, let \(a = x+y\), \(b=y+z\) and \(c=z+x\). Note that \(x\), \(y\) and \(z\) are positive reals.

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Ummm... any hint?

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Using \(Titu's..lemma\) we get \[\sum_{cyc}\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\geq\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\] By using \(RMS-AM\) inequality we get \[\sqrt{\frac{2b^{2}+2c^{2}-a^{2}+2c^{2}+2a^{2}-b^{2}+2b^{2}+2a^{2}-c^{2}}{3}}\geq\frac{\sqrt{2b^{2}+2c^{2}-a^{2}}}{3}\] This implies \[3\sqrt{a^{2}+b^{2}+c^{2}}\geq \sqrt{2b^{2}+2c^{2}-a^{2}}\] \[\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\geq\frac{ (\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}{3\sqrt{a^{2}+b^{2}+c^{2}}}\geq\frac{ (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{\sqrt{a^{2}+b^{2}+c^{2}}}\] Therefore \[\sum_{cyc}\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\geq\frac{(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{\sqrt{a^{2}+b^{2}+c^{2}}}\] Now anyone please help me to finish the solution.

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

In using \( 3 \sqrt{ a^2 + b^2 + c^2 } \geq \sqrt{ 2 b^2 + 2c^2 - a ^2 } \) you made it way too loose, which is why the bound does not work. For example, we have\( \sqrt{2} \sqrt{ a^2 + b^2 + c^2 } \geq \sqrt{2 a^2 + 2b^2 + 2c^2 } > \sqrt{ 2 b^2 + 2c^2 - a^2} \).

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

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