Sa se arate ca daca \( ABC \) este un triunghi ascutitunghic, atunci
\( \sum\ \frac {b+c}{a}\ \ge\ \frac {(R+r)(R+2r)}{Rr}\ \ge\ 3\left(1+\sqrt[3]{\frac {R}{2r}}\right). \)
Inegalitate intr-un triunghi ascutitunghic.
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Virgil Nicula
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opincariumihai
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Obtinem succesiv :
\( \sum \frac{b+c}{a}= \frac{ \sum a \sum bc}{abc}-3=\frac{p^2+r^2-2Rr}{2Rr} \)
Folosind in continuare inegalitatea \( p^2\ge 2R^2+8Rr+3r^2 \) care se gaseste aici vom obtine
\( \sum \frac{b+c}{a}\geq \frac{R^2+2r^2+3rR}{Rr}=\frac {(R+r)(R+2r)}{Rr}. \)
Pentru partea a doua, notand cu \( x^3=R/2r \) inegalitatea \( \frac {(R+r)(R+2r)}{Rr}\ \ge\ 3\left(1+\sqrt[3]{\frac {R}{2r}}\right) \) se scrie echivalent
\( (1+2x^3)(1+\frac{1}{x^3})\ \ge\ 3(1+x) \)
echivalent cu
\( 2x^3+\frac{1}{x^3} \geq 3x \)
inegalitate care reiese usor din medii.
\( \sum \frac{b+c}{a}= \frac{ \sum a \sum bc}{abc}-3=\frac{p^2+r^2-2Rr}{2Rr} \)
Folosind in continuare inegalitatea \( p^2\ge 2R^2+8Rr+3r^2 \) care se gaseste aici vom obtine
\( \sum \frac{b+c}{a}\geq \frac{R^2+2r^2+3rR}{Rr}=\frac {(R+r)(R+2r)}{Rr}. \)
Pentru partea a doua, notand cu \( x^3=R/2r \) inegalitatea \( \frac {(R+r)(R+2r)}{Rr}\ \ge\ 3\left(1+\sqrt[3]{\frac {R}{2r}}\right) \) se scrie echivalent
\( (1+2x^3)(1+\frac{1}{x^3})\ \ge\ 3(1+x) \)
echivalent cu
\( 2x^3+\frac{1}{x^3} \geq 3x \)
inegalitate care reiese usor din medii.