Sa se arate ca in orice triunghi are loc inegalitatea:\( \frac{sqrt{r_br_c}}{a}+\frac{sqrt{r_cr_a}}{b}+\frac{sqrt{r_ar_b}}{c}\geq\frac{3\sqrt{3}}{2}. \)
Nica Nicolae
Inegalitate geometrica
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Inegalitate geometrica
"Matematica este asemeni constitutiei unei tari, ale carei legi sunt: leme, teoreme, definitii..." Nica Nicolae
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Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
\( r_a=\frac{S}{p-a} \) si analoagele inegalitatea devine
\( \sum \frac{\sqrt{p(p-a)}}{a}\ge\frac{3\sqrt{3}}{2} \) sau
\( \sum\frac{\sqrt{x}}{y+z}\ge\frac{3\sqrt{3}}{2\sqrt{x+y+z}} \)
sau notand x+y+z=1
\( \sum\frac{\sqrt{x}}{1-x}\ge\frac{3\sqrt{3}}{2} \) care este adevarata deoarece
\( \sqrt{x}(1-x)\le\frac{2}{3\sqrt{3}} \) si analoagele.
\( \sum \frac{\sqrt{p(p-a)}}{a}\ge\frac{3\sqrt{3}}{2} \) sau
\( \sum\frac{\sqrt{x}}{y+z}\ge\frac{3\sqrt{3}}{2\sqrt{x+y+z}} \)
sau notand x+y+z=1
\( \sum\frac{\sqrt{x}}{1-x}\ge\frac{3\sqrt{3}}{2} \) care este adevarata deoarece
\( \sqrt{x}(1-x)\le\frac{2}{3\sqrt{3}} \) si analoagele.
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Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Se arata usor ca \( \sum \frac{sqrt{r_br_c}}{a}\geq\frac{3\sqrt{3}}{2}\ \Longleftrightarrow\ \sum \frac{\sqrt{p-a}}{a}\ge\frac{3\sqrt{3}}{2\sqrt p} \) . Aplicam inegalitatea mediilor :
\( \frac a2\cdot\frac a2\cdot (p-a)\le\left[\frac {\frac a2+\frac a2+(p-a)}{3}\right]^3\Longrightarrow27a^2(p-a)\le 4p^3\Longleftrightarrow 3a\sqrt {3(p-a)}\le 2p\sqrt p\Longrightarrow \)
\( \frac {1}{a\sqrt {p-a}}\ \ge\ \frac {3\sqrt 3}{2p\sqrt p}\ (*)\ . \) Asadar, \( \sum \frac{\sqrt{p-a}}{a}=\sum\frac {p-a}{a\sqrt {p-a}}\ \stackrel {(*)}{\ge}\ \frac {3\sqrt 3}{2p\sqrt p}\sum (p-a)=\frac {3\sqrt 3}{2\sqrt p}\ . \)
Aplicatii. Sa se arate ca :
\( 1.\ \ \sqrt {r_a}+\sqrt {r_b}+\sqrt {r_c}\ge3\sqrt {3r}\ . \)
\( 2.\ \ \sum\sqrt[3]{a(b+c-a)^2}\ge a+b+c\ \Longleftrightarrow\ \sum\sqrt[3]{\frac {a}{r_a^2}}\ge \sqrt[3]{\frac{a+b+c}{r^2}}\ . \)
\( 3.\ \ \sum\sqrt[3]{bc(b+c-a)}\ \le \ a+b+c\ . \)
\( \frac a2\cdot\frac a2\cdot (p-a)\le\left[\frac {\frac a2+\frac a2+(p-a)}{3}\right]^3\Longrightarrow27a^2(p-a)\le 4p^3\Longleftrightarrow 3a\sqrt {3(p-a)}\le 2p\sqrt p\Longrightarrow \)
\( \frac {1}{a\sqrt {p-a}}\ \ge\ \frac {3\sqrt 3}{2p\sqrt p}\ (*)\ . \) Asadar, \( \sum \frac{\sqrt{p-a}}{a}=\sum\frac {p-a}{a\sqrt {p-a}}\ \stackrel {(*)}{\ge}\ \frac {3\sqrt 3}{2p\sqrt p}\sum (p-a)=\frac {3\sqrt 3}{2\sqrt p}\ . \)
Aplicatii. Sa se arate ca :
\( 1.\ \ \sqrt {r_a}+\sqrt {r_b}+\sqrt {r_c}\ge3\sqrt {3r}\ . \)
\( 2.\ \ \sum\sqrt[3]{a(b+c-a)^2}\ge a+b+c\ \Longleftrightarrow\ \sum\sqrt[3]{\frac {a}{r_a^2}}\ge \sqrt[3]{\frac{a+b+c}{r^2}}\ . \)
\( 3.\ \ \sum\sqrt[3]{bc(b+c-a)}\ \le \ a+b+c\ . \)