[Toán 9] Bất đẳng thức hình học

H

huytrandinh

ta có
[TEX]S_{ABC}=\sqrt{p(p-a)(p-b)(p-c)}[/TEX]
[TEX]=\frac{1}{2\sqrt{2}}\sqrt{p(a+b-c)(a+c-b)(b+c-a)}[/TEX]
[TEX].\sqrt{(a+b-c)(b+c-a)}\leq b[/TEX]
[TEX].\sqrt{(a+b-c)(a+c-b)}\leq a[/TEX]
[TEX].\sqrt{(a+c-b)(b+c-a)}\leq c[/TEX]
[TEX]=>(a+b-c)(a+c-b)(b+c-a)\leq abc[/TEX]
[TEX]<=>\sqrt{p(a+b-c)(a+c-b)(b+c-a)}\leq \sqrt{p.abc}[/TEX]
[TEX]=>S_{ABC}= \frac{abc}{4R}\leq \frac{1}{2\sqrt{2}}\sqrt{p.abc}[/TEX]
[TEX]<=>p\geq \frac{abc}{2}<=>a+b+c\geq abc[/TEX]
 
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