If a, b, c are three sides of the $$ riangle ABC$$ and A, B, C are their corresponding vertex, show that
$$a^2 an\dfrac{A}{2}+b^2 an\dfrac{B}
{2}+c^2 an\dfrac{C}{2}\geq 4S $$
in which S is the area of the triangle.

Question

If a, b, c are three sides of the $$	riangle ABC$$ and A, B, C are their corresponding vertex, show that 
$$a^2	an\dfrac{A}{2}+b^2	an\dfrac{B}
{2}+c^2	an\dfrac{C}{2}\geq  4S      $$
in which S is the area of the triangle.

Uniteasy Admin · Accepted Answer

Using half angle identity and the law of cosines 
$$\cos^2\dfrac{A}{2}=\dfrac{1+\cos A}{2}  $$
$$=\dfrac{1+\dfrac{b^2+c^2-a^2}{2bc} }{2} $$
$$=\dfrac{(b+c)^2-a^2}{4bc} $$
$$=\dfrac{(b+c+a)(b+c-a)}{4bc} $$
 $$\dfrac{a^2	an\dfrac{A}{2}}{S} =\dfrac{a^2\cdotp \dfrac{\sin \dfrac{A}{2} }{\cos \dfrac{A}{2} }  }{\dfrac{1}{2}bc\sin A } 
=\dfrac{a^2}{bc\cos^2\dfrac{A}{2}  } $$
$$=\dfrac{4a^2}{(b+c+a)(b+c-a)} $$
Similarly, we can get,
$$\dfrac{b^2	an\dfrac{B}{2} }{S}= \dfrac{4b^2}{(a+b+c)(a+c-b)} $$
$$\dfrac{c^2\cos\dfrac{C}{2} }{S}=\dfrac{4c^2}{(a+b+c)(a+b-c)}  $$
On the other hand, the following inequality can be obtained by using Cauch Inequality.
$$(\dfrac{a^2}{b+c-a}+\dfrac{b^2}{a+c-b}+\dfrac{c^2}{a+b-c}) [(b+c-a)+(a+c-b)+(a+b-c)]\geq (a+b+c)^2 $$
$$ \dfrac{a^2}{b+c-a}+\dfrac{b^2}{a+c-b}+\dfrac{c^2}{a+b-c} \geq a+b+c $$
$$\dfrac{a^2	an\dfrac{A}{2}}{S}+\dfrac{b^2	an\dfrac{B}{2} }{S}+\dfrac{c^2\cos\dfrac{C}{2} }{S}$$
$$=(a+b+c)( \dfrac{a^2}{b+c-a}+\dfrac{b^2}{a+c-b}+\dfrac{c^2}{a+b-c})\geq 4 $$
Therefore,
$$a^2	an\dfrac{A}{2}+b^2	an\dfrac{B}
{2}+c^2	an\dfrac{C}{2}\geq 4S   $$

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