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# 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
3 years, 11 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. · 3 years, 11 months ago

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 $$x \to 0$$. · 3 years, 11 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. · 3 years, 11 months ago