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.

Note by Đức Việt Lê
4 years, 9 months ago

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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 - 4 years, 9 months ago

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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 - 4 years, 9 months ago

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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 - 4 years, 9 months ago

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