A geometric inequality

It's hard for me to try to prove this inequation. Can anyone do?

Let triangle ABC. a = BC , b = AC, c = AB. Let A, B, C are angle measurements of that triangle. Prove that

$ 60 \leq \frac{aA + bB + cC}{a + b + c} < 90 $

That's the problem. But are there a maximum value of $\frac{aA + bB + cC}{a + b + c}$ ? That's the thing I'm still wondering too.

Thank you.

#Geometry #HelpMe! #MathProblem #Math

Easy Math Editor

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Comments

First, we prove the LHS inequality.

Without loss of generality, assume that $A \geq B \geq C$ . Then obviously $a \geq b \geq c$ . Using the Chebyshev's inequality, which you can find here: http://en.wikipedia.org/wiki/Chebyshev's_inequality , we get: $aA + bB + cC \geq \frac {(a+b+c)(A+B+C)}{3} = 60(a+b+c)$ which leads to the LHS inequality.

Now we move to RHS inequality.

It is easy to see that $a,b,c < \frac {a+b+c}{2}$ respectively. Then $\frac{aA+bB+cC}{a+b+c} < \frac{A+B+C}{2} = 90$ Notice: You should express those measurements in radian instead of degree next time.

Linh Tran - 7 years, 9 months ago

Adding to Linh T.'s post...

You cannot achieve $90$ , but you can get as close to $90$ as you like. If $ABC$ is isosceles with $A = B = x^\circ$ , then $C = (180-2x)^\circ$ and we have $a=b$ and $c = 2a\cos x^\circ$ . Then $\begin{array}{rcl} \frac{aA+bB+cC}{a+b+c} & = & \frac{ax + ax + 2a(180-2x)\cos x^\circ}{a + a + 2a\cos x^\circ} \\ & = & \frac{2a[x + (180-2x)\cos x^\circ]}{2a(1 + \cos x^\circ)} \; = \; \frac{x + (180-2x)\cos x^\circ}{1 + \cos x^\circ} \end{array}$ and this tends to $90$ as $x \to 0$ .

Mark Hennings - 7 years, 9 months ago

Thank you for noticing. I will use $\frac{\pi}{3}$ and $\frac{\pi}{2}$ instead, on the next time if i meet those numbers again.

Đức Việt Lê - 7 years, 9 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}`	$\boxed{123}$