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

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

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

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

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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. \(\square\) – Alan Yan · 1 year, 1 month ago

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– Calvin Lin Staff · 1 year, 1 month ago

How would one think of coming up with that first equation?Log in to reply

http://yufeizhao.com/wc08/ineq.pdf – Alan Yan · 1 year, 1 month ago

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– Shyambhu Mukherjee · 1 year, 1 month ago

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

– Lakshya Sinha · 1 year, 1 month ago

I am not able to prove it can you please helpLog in to reply

– Alan Yan · 1 year, 1 month ago

Just expand and simplify the expression I wrote.Log in to reply

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. – Sharky Kesa · 1 year, 1 month ago

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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? – Figel Ilham · 1 year, 1 month ago

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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. – Shivam Jadhav · 1 year, 1 month ago

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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. – Calvin Lin Staff · 1 year, 1 month ago

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– Shivam Jadhav · 1 year, 1 month ago

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

– Shyambhu Mukherjee · 1 year, 1 month ago

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.Log in to reply

– Alan Yan · 1 year, 1 month ago

If I have not made any computational errors, the measure of the A-median would be \(m_a = \frac{\sqrt{2b^2 + 2c^2 - a^2}}{2} \). Your proposal is to prove that \[\frac{\sqrt{2b^2 + 2c^2 - a^2}}{2} \leq \frac{\sqrt{3}a}{2} \implies 2b^2 + 2c^2 - a^2 \leq 3a^2 \implies b^2 + c^2 \leq 2a^2\] which is obviously false given a triangle such that \(b , c > a\).Log in to reply

– Shyambhu Mukherjee · 1 year, 1 month ago

Thanks . actually I guessed it only. Bad guess I guess.Log in to reply

– Shivam Jadhav · 1 year, 1 month ago

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

– Reynan Henry · 1 year, 1 month ago

Is the degenerate triangle excluded in the problem?Log in to reply

In this case, I don't think it matters. The inequality is true for \( \{ a, b, c \} = \{ 0, 1, 1 \} \). – Calvin Lin Staff · 1 year, 1 month ago

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– Lakshya Sinha · 1 year, 1 month ago

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 emailLog in to reply

change your password which will log you out of all other devices. – Calvin Lin Staff · 1 year, 1 month ago

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Ummm... any hint? – Lakshya Sinha · 1 year, 1 month ago

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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. – Shivam Jadhav · 1 year, 1 month ago

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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\( \sqrt{2} \sqrt{ a^2 + b^2 + c^2 } \geq \sqrt{2 a^2 + 2b^2 + 2c^2 } > \sqrt{ 2 b^2 + 2c^2 - a^2} \). – Calvin Lin Staff · 1 year, 1 month ago

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– Pranjal Mittal · 1 year, 1 month ago

Sir, please can you explain me the "bound" thing??Log in to reply