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January 2nd, 2019, 12:50 PM   #1
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Inequality problem

Let triangle with side a,b,c such that a+b+c=10.
Prove this
in.png
I need help with this problem by use of triangle inequality, AG or similar inequality that I know. So I ask for hint.

Last edited by skipjack; January 3rd, 2019 at 06:42 AM.
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January 2nd, 2019, 01:15 PM   #2
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Quote:
Originally Posted by jsf View Post
Let triangle with side a,b,c such that a+b+c=10.
Prove this
Attachment 10106
I need help with this problem by use of triangle inequality, AG or similar inequality that I know. So I ask for hint.
I can't see the image. (I can manually type it in and view it on the web but that's all.)

So:
$\displaystyle a + b + c = 10$

Here's the inequality in the image:
$\displaystyle a^2 + b^2 + c^2 + \frac{2}{5} abc < 50$

Can we take it that you have to derive the inequality from the first equation?

-Dan

Last edited by skipjack; January 3rd, 2019 at 06:44 AM.
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January 2nd, 2019, 06:04 PM   #3
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I've posted the image in-line.
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January 2nd, 2019, 06:23 PM   #4
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Because the variables in the inequality can be interchanged with respect to value and position the equation has cyclic symmetry and an extremum is reached when all three sides are equal. After you come up with a value for that all you have to do is devise another triangle to show that the LHS of the inequality is less than the previously calculated value.
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January 3rd, 2019, 08:03 AM   #5
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Quote:
Originally Posted by jsf View Post
I ask for hint.
Hint: as $a + b + c = 10$ and $b + c > a$, $a < 5$.
Similarly, $b < 5$ and $c < 5$, so $0 < (5 - a)(5 - b)(5 - c)$.
Multiply that inequality by 2/5 and use the fact that $(a + b + c)^2 = 100$.
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