# 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. Note by Sharky Kesa
5 years, 7 months ago

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$\textbf{Solution.}$ 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. $\square$

- 5 years, 7 months ago

- 5 years, 7 months ago

Just expand and simplify the expression I wrote.

- 5 years, 7 months ago

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

- 5 years, 7 months ago

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

Staff - 5 years, 7 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

- 5 years, 7 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 $\frac{3}{2}\leq\sum_{cyc}\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}$.

- 5 years, 7 months ago

Is the degenerate triangle excluded in the problem?

- 5 years, 7 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 $\{ a, b, c \} = \{ 0, 1, 1 \}$.

Staff - 5 years, 7 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 email

- 5 years, 7 months ago

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

Staff - 5 years, 7 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.

Staff - 5 years, 7 months ago

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

- 5 years, 7 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.

- 5 years, 7 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 $b , c > a$.

- 5 years, 7 months ago

Thanks . actually I guessed it only. Bad guess I guess.

- 5 years, 7 months ago

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

- 5 years, 7 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?

- 5 years, 7 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.

- 5 years, 7 months ago

Ummm... any hint?

- 5 years, 7 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.

- 5 years, 7 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$\sqrt{2} \sqrt{ a^2 + b^2 + c^2 } \geq \sqrt{2 a^2 + 2b^2 + 2c^2 } > \sqrt{ 2 b^2 + 2c^2 - a^2}$.

Staff - 5 years, 7 months ago

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

- 5 years, 7 months ago