BDT tam giác

S

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$\sqrt{a^2-(b-c)^2}+\sqrt{b^2-(c-a)^2}+\sqrt{c^2-\sqrt{a-b}^2}=\sqrt{(a+b-c)(a+c-b)}+\sqrt{(b+c-a)(a+b-c)}+\sqrt{(c+a-b)(c+b-a)}$

Đặt $a+b-c=x;b+c-a=y;a+c-b=z$

$\leftrightarrow a=\dfrac{x+z}{2};b=\dfrac{x+y}{2};c=\dfrac{y+z}{2}$

Bất đẳng tương được với

$\sqrt{xy}+\sqrt{yz}+\sqrt{xz} \leq 0,5(\sqrt{(x+y)(y+z)}+\sqrt{(x+y)
(x+z)}+\sqrt{(x+z)(y+z)}$

Ta có: $\sqrt{xy}+\sqrt{yz} \leq \sqrt{(x+y)(y+z)}$

Ta có: $\sqrt{xy}+\sqrt{xz} \leq \sqrt{(x+y)(x+z)}$

Ta có: $\sqrt{yz}+\sqrt{xz} \leq \sqrt{(x+z)(y+z)}$

Cộng từng vế $\leftrightarrow dpcm \leftrightarrow x=y=z$


 
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