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Suppose $a, b, c$ are the lengths of three triangular edges. Prove that:

$$\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$$

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5  
Hint: Multiply with $\sqrt{a+b+c}$ to obtain $4A$ on the left hand side (Heron's formula). –  Hagen von Eitzen Dec 20 '12 at 12:58

2 Answers 2

up vote 1 down vote accepted

As the hint give in the comment says (I denote by $S$ the area of $ABC$ and by $R$ the radius of its circumcircle), if you multiply your inequality by $\sqrt{a+b+c}$ you'll get

$$4S \leq \frac{3\sqrt{3}abc}{a+b+c}$$

which is eqivalent to

$$a+b+c \leq 3\sqrt{3}\frac{abc}{4S}=3\sqrt{3}R.$$


This inequality is quite known. If you want a proof, you can write $a=2R \sin A$ (and the other two equalities) and get the equivalent inequality

$$ \sin A +\sin B +\sin C \leq \frac{3\sqrt{3}}{2}$$

which is an easy application of the Jensen inequality for the concave function $\sin : [0,\pi] \to [0,1]$.

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Thank you very much –  H.T.H Dec 21 '12 at 11:14

Here is another way to proof $ sinA+sinB+sinC\leq \dfrac{3\sqrt{3}}{2}$ in case some one don't know the Jensen inequality.

in general ,let $A \geq B \geq C$,then $C< \dfrac{\pi}{2}$ and $cos\dfrac{C}{2}>0$

$ sinA+sinB+sinC=2sin\dfrac{A+B}{2}cos\dfrac{A-B}{2}+2sin\dfrac{C}{2}cos\dfrac{C}{2}=2cos\dfrac{C}{2}(cos\dfrac{A-B}{2}+sin\dfrac{C}{2}) \leq 2cos\dfrac{C}{2}(1+sin\dfrac{C}{2})$ ,

since $cos\dfrac{A-B}{2} \leq 1$,when $A=B$, it equals 1.

from $x_1*x_2 \leq (\dfrac{x_1+x_2}{2})^2$, we can get $x_1x_2x_3x_4 \leq (\dfrac{x_1+x_2+x_3+x_4}{4})^4$

$ 2cos\dfrac{C}{2}(1+sin\dfrac{C}{2})$$=2\sqrt{1-sin^2\dfrac{C}{2}}(1+sin\dfrac{C}{2})$$=2\sqrt{(1-sin^2\dfrac{C}{2})(1+sin\dfrac{C}{2})^2}=2\sqrt{(1-sin\dfrac{C}{2})(1+sin\dfrac{C}{2})(1+sin\dfrac{C}{2})(1+sin\dfrac{C}{2})}$$=2\sqrt{\dfrac{1}{3}(3-3sin\dfrac{C}{2})(1+sin\dfrac{C}{2})(1+sin\dfrac{C}{2})(1+sin\dfrac{C}{2})}$ $ \leq 2\sqrt{\dfrac{1}{3}(\dfrac{3-3sin\dfrac{C}{2}+1+sin\dfrac{C}{2}+1+sin\dfrac{C}{2}+1+sin\dfrac{C}{2}}{4})^4}$=$2\sqrt{\dfrac{1}{3}(\dfrac{6}{4})^4}$=$2\sqrt{\dfrac{1}{3}(\dfrac{3}{2})^4}=$$\dfrac{3\sqrt{3}}{2} $ when and only when $3-3sin\dfrac{C}{2}=1+sin\dfrac{C}{2}$,ie $sin\dfrac{C}{2}=\dfrac{1}{2}$,you get the equals. put $A=B$ together, we get when and only when $A=B=C$, the LHS=$\dfrac{3\sqrt{3}}{2} $.

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