Bài 3 nha mọi người tks nhìu

Question

Cho

$a,b,c$ là độ dài các cạnh của một tam giác. Chứng minh rằng:

$\sqrt{a^{2}-(b-c)^{2}}+\sqrt{b^{2}-(c-a)^{2}}+\sqrt{c^{2}-(a-b)^{2}}\leq\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq a+b+c$

Pooh · Answer 1 · 8/22/2013 3:36:57 PM

2;ta co

$xy+yz+xz\leq x^2+y^2+z^2$ (luon dung ) nen ta co dieu phai CM

Trả lời 22-08-13 03:36 PM

Pooh
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Pooh · Answer 2 · 8/22/2013 3:35:29 PM

ta co

$a^2-(b-c)^2=(a+b-c)(a-b+c)$ a+b-c=x>0

$b^2-(a-c)^2=(a+b-c)(-a+b+c)$ Dat a-b+c=y>0

$c^2-(a-b)^2=(a-b+c)(-a+b+c)$ -a+b+c+z>0

bat dang thuc ban dau tuong duong

$\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\leq$

$\sqrt{(\frac{x+y}2)(\frac{x+z}2)}+\sqrt{(\frac{x+z}2)(\frac{y+z}2)}$

$+\sqrt{(\frac{x+y}2)(\frac{x+z}2)}$

ma

$\sqrt{(x+z)(x+y)}\geq\sqrt{(\sqrt{xy}+\sqrt{xz})^2}=\sqrt{xy}+\sqrt{xz}$

$\sqrt{(y+z)(x+y)}\geq\sqrt{xy}+\sqrt{yz}$

$\sqrt{(x+z)(y+z)}\geq\sqrt{xz}+\sqrt{yz}$

cong ba BDT tren ta co dieu phai CM

2 Đáp án