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

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

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

#Algebra #Sharky

Easy Math Editor

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Comments

$\textbf{Solution.}$ 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 - 5 years, 6 months ago

I am not able to prove it can you please help

Department 8 - 5 years, 6 months ago

Just expand and simplify the expression I wrote.

Alan Yan - 5 years, 6 months ago

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

Shyambhu Mukherjee - 5 years, 6 months ago

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

Calvin Lin Staff - 5 years, 6 months ago

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

Alan Yan - 5 years, 6 months ago

$cosA=\frac{b^{2}+c^{2}-a^{2}}{2bc} (cosine..rule)$
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 - 5 years, 6 months ago

Is the degenerate triangle excluded in the problem?

Reynan Henry - 5 years, 6 months ago

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

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

Calvin Lin Staff - 5 years, 6 months ago

@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 - 5 years, 6 months ago

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

Calvin Lin Staff - 5 years, 6 months ago

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.

Calvin Lin Staff - 5 years, 6 months ago

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

Shivam Jadhav - 5 years, 6 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.

Shyambhu Mukherjee - 5 years, 6 months ago

@Shyambhu Mukherjee – 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 $b , c > a$ .

Alan Yan - 5 years, 6 months ago

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

Shyambhu Mukherjee - 5 years, 6 months ago

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

Shivam Jadhav - 5 years, 6 months ago

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 - 5 years, 6 months ago

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 - 5 years, 6 months ago

Ummm... any hint?

Department 8 - 5 years, 6 months ago

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 - 5 years, 6 months ago

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 $\sqrt{2} \sqrt{ a^2 + b^2 + c^2 } \geq \sqrt{2 a^2 + 2b^2 + 2c^2 } > \sqrt{ 2 b^2 + 2c^2 - a^2}$ .

Calvin Lin Staff - 5 years, 6 months ago

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

Pranjal Mittal - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$