Točno
1. lipnja 2014. 12:52 (10 godine, 1 mjesec)
Neka su
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
,
![b](/media/m/e/e/c/eec0d7323095a1f2101fc1a74d069df6.png)
i
![c](/media/m/e/a/3/ea344283b6fa26e4a02989dd1fb52a51.png)
duljine stranica trokuta opsega
![1](/media/m/a/9/1/a913f49384c0227c8ea296a725bfc987.png)
. Dokaži da vrijedi
%V0
Neka su $a$, $b$ i $c$ duljine stranica trokuta opsega $1$. Dokaži da vrijedi $$\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} < 1 + \frac {\sqrt{2}}{2}.$$
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Uvedimo supstituciju
![(a,b,c)=(x+y,y+z,z+x)](/media/m/b/f/3/bf374b0bbe190bf6dc88ada8637ebc09.png)
, koju možemo napraviti jer su to stranice trokuta (
![x](/media/m/f/1/8/f185adeed9bd346bc960bca0147d7aae.png)
,
![y](/media/m/c/c/0/cc082a07a517ebbe9b72fd580832a939.png)
i
![z](/media/m/d/2/4/d241a79f1fdd0ce9a8f3f91570ba5d62.png)
su duljine dužina između vrhova trokuta i dirališta upisane kružnice).
Zadani uvjet postaje
![x+y+z=\frac{1}{2}](/media/m/c/5/8/c581df2130f81840556dfb5bc3639a10.png)
.
![\sum{\sqrt{(x+y)^2+(z+y)^2}} < 1 + \frac{\sqrt{2}}{2}](/media/m/c/a/b/cab6b3f29d720b76dc4919c4de881e21.png)
Uzmimo vektore
![\mathbf{A}=(x,z)](/media/m/d/8/9/d8981d70f22569c622b0bdbdd9cfe80c.png)
i
![\mathbf{B}=(y,y)](/media/m/f/5/9/f59bdc19a22a39d0fbef73b3a984bc05.png)
.
Po nejednakosti trokuta vrijedi
![|\mathbf{A}+\mathbf{B}| \le |\mathbf{A}| + |\mathbf{B}|](/media/m/5/1/8/5185a58232e8f4489ef5c5759cf1d705.png)
.
![\mathbf{A} + \mathbf{B} = (x+y, z+y)](/media/m/4/f/1/4f138258fedea38680869d28ffbd6c63.png)
![\sqrt{(x+y)^2 + (z+y)^2} \le \sqrt{x^2+z^2} + \sqrt{y^2+y^2} = \sqrt{x^2 + z^2} + \sqrt{2}y](/media/m/4/b/7/4b777cede1db663a22b3da1b4fad8759.png)
Koristeći trivijalnu nejednakost
![\sqrt{x^2+z^2} < x+z](/media/m/2/7/9/2792338e431639f9a32eb20cb193a527.png)
, dobivamo:
![\sqrt{(x+y)^2 + (z+y)^2} < x+z+\sqrt{2}y](/media/m/2/2/d/22de017519ced711c7a57b3013c8802d.png)
Zbrajajući tri takve nejednakosti, dobivamo:
%V0
Uvedimo supstituciju $(a,b,c)=(x+y,y+z,z+x)$, koju možemo napraviti jer su to stranice trokuta ($x$, $y$ i $z$ su duljine dužina između vrhova trokuta i dirališta upisane kružnice).
Zadani uvjet postaje $x+y+z=\frac{1}{2}$.
$ \sum{\sqrt{(x+y)^2+(z+y)^2}} < 1 + \frac{\sqrt{2}}{2}$
Uzmimo vektore $\mathbf{A}=(x,z)$ i $\mathbf{B}=(y,y)$.
Po nejednakosti trokuta vrijedi $|\mathbf{A}+\mathbf{B}| \le |\mathbf{A}| + |\mathbf{B}|$.
$ \mathbf{A} + \mathbf{B} = (x+y, z+y)$
$ \sqrt{(x+y)^2 + (z+y)^2} \le \sqrt{x^2+z^2} + \sqrt{y^2+y^2} = \sqrt{x^2 + z^2} + \sqrt{2}y $
Koristeći trivijalnu nejednakost $\sqrt{x^2+z^2} < x+z$, dobivamo:
$ \sqrt{(x+y)^2 + (z+y)^2} < x+z+\sqrt{2}y $
Zbrajajući tri takve nejednakosti, dobivamo:
$ \sum{\sqrt{(x+y)^2+(z+y)^2}} < \sum{x+z+\sqrt{2}y} = (2+\sqrt{2})(x+y+z) = 1 + \frac{\sqrt{2}}{2}$ $\blacksquare$
26. travnja 2015. 19:36 | ikicic | Točno |