Točno
15. lipnja 2015. 01:34 (9 godine, 1 mjesec)
Neka su
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
i
![b](/media/m/e/e/c/eec0d7323095a1f2101fc1a74d069df6.png)
duljine kateta, a
![c](/media/m/e/a/3/ea344283b6fa26e4a02989dd1fb52a51.png)
duljina hipotenuze pravokutnog trokuta.
Dokaži da vrijedi
%V0
Neka su $a$ i $b$ duljine kateta, a $c$ duljina hipotenuze pravokutnog trokuta.
Dokaži da vrijedi $$\left(1+\frac ca\right)\left(1+\frac cb\right) \geq 3+2\sqrt{2}.$$
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%V0
$(1+\frac{c}{a})(1+\frac{c}{b})\geq3+2\sqrt2$
$\frac{c}{a}+\frac{c}{b}+\frac{c^2}{ab}\geq2+2\sqrt2$
$\frac{c^2}{ab} = \frac{a^2+b^2}{ab}$
Prema $A \geq G$:
$\frac{a^2+b^2}{2}\geq\sqrt{a^2b^2}$
$\frac{a^2+b^2}{2}\geq ab$
$\frac{a^2+b^2}{ab}\geq 2$
Prema $A \geq G$:
$\frac{\frac{c}{a}+\frac{c}{b}}{2}\geq\sqrt{\frac{c}{a}\cdot\frac{c}{b}}$
$(\frac{\frac{c}{a}+\frac{c}{b}}{2})^2\geq\frac{c^2}{ab}$
$(\frac{\frac{c}{a}+\frac{c}{b}}{2})^2\geq\frac{a^2+b^2}{ab}$
Prema tranzitivnosti:
$(\frac{\frac{c}{a}+\frac{c}{b}}{2})^2\geq2$
$\frac{c}{a}+\frac{c}{b}\geq2\sqrt2$
Zbrajajući nejednakosti:
$\frac{c}{a}+\frac{c}{b}+\frac{a^2+b^2}{ab}\geq2+2\sqrt2$
$\frac{c}{a}+\frac{c}{b}+\frac{c^2}{ab}\geq2+2\sqrt2$
$Q.E.D.$
15. lipnja 2015. 08:40 | ikicic | Točno |