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

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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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, 4 months ago

Log in to reply

– Calvin Lin Staff · 1 year, 4 months 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, 4 months ago

Log in to reply

– Shyambhu Mukherjee · 1 year, 4 months 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, 4 months ago

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

– Alan Yan · 1 year, 4 months 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, 4 months ago

Log in to reply

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, 4 months ago

Log in to reply

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

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, 4 months ago

Log in to reply

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, 4 months ago

Log in to reply

– Shivam Jadhav · 1 year, 4 months ago

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

– Shyambhu Mukherjee · 1 year, 4 months 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, 4 months 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, 4 months ago

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

– Shivam Jadhav · 1 year, 4 months 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, 4 months 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, 4 months ago

Log in to reply

– Lakshya Sinha · 1 year, 4 months 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, 4 months ago

You canLog in to reply

Ummm... any hint? – Lakshya Sinha · 1 year, 4 months ago

Log in to reply

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, 4 months ago

Log in to reply

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, 4 months ago

Log in to reply

– Pranjal Mittal · 1 year, 4 months ago

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