Inegalitate in triunghi

[Mateescu Constantin]

\( \triangle ABC\ \Longrightarrow\ \frac 1{b+c}+\frac 1{c+a}+\frac 1{a+b}\ \le\ \frac{3(R+r)}{4S} \) .


Virgil Nicula

[Marius Mainea]

Folosind inegalitatile Gerretsen (\( p^2\le 4R^2+4Rr+3r^2 \)) si Euler (\( R\ge 2r \))

\( LHS\le \sum\frac{1}{2\sqrt{ab}}\le\frac{1}{2}\sum\frac{1}{a}=\frac{1}{2}\frac{p^2+r^2+4Rr}{abc}\le\frac{3(R+r)}{4\frac{abc}{4R}}=RHS \)

[Virgil Nicula]

\( \frac {9}{4p}\stackrel{(CBS)}{\le}\sum\frac {1}{b+c}\le\sum\frac {b+c}{4bc}=\sum\frac {a(b+c)}{4abc}=\frac {\sum bc}{2abc}=\frac {p^2+r^2+4Rr}{8RS}\ \stackrel{(\mathrm{Gerretsen})}{\le}\frac {\left(4R^2+4Rr+3r^2\right)+r^2+4Rr}{8RS}= \)

\( \frac {(R+r)\cdot (R+r)}{2RS}\le\frac {\frac {3R}{2}\cdot (R+r)}{2RS}=\frac {3(R+r)}{4S}\ \Longleftrightarrow\ \underline{\overline{\left\|\ \frac {9}{4p}\ \le\ \sum\frac {1}{b+c}\ \le\ \frac {3(R+r)}{4S}\ \right\|}} \) . Retineti echivalenta

\( \underline{\overline{\left\|\ 4r(5R-r)\ \le\ ab+bc+ca\ \le\ 4(R+r)^2\ \right\|}}\ \Longleftrightarrow\ \overline{\underline{\left\|\ 16Rr-5r^2\ \le\ p^2\ \le\ 4R^2+4Rr+3r^2\ \right\|}} \) (Gerretsen).

Aplicatie. Sa se arate ca \( \left(\frac {p}{R+r}\right)^2\ \le\ \sum\frac {a^2}{bc}\ \le\ \frac {2R-r}{r} \) si \( \sum \frac {1}{a(b+c)}\ \ge \frac {9}{8(R+r)^2} \) .

[Mateescu Constantin]

Sa se arate ca \( \left(\frac {p}{R+r}\right)^2\ \le\ \sum\frac {a^2}{bc}\ \le\ \frac {2R-r}{r} \) si \( \sum\frac{1}{a(b+c)}\ge \frac{9}{8(R+r)^2} \) .

Demonstratie.

\( \left\(\frac{p}{R+r}\right\)^2=\frac{4p^2}{(4R^2+4Rr+3r^2)+r^2+4Rr}\ \stackrel{\mbox{(Gerretsen)}}{\le}\ \frac{4p^2}{p^2+r^2+4Rr}=\frac{(a+b+c)^2}{ab+bc+ca}\ \stackrel{\mbox{(C.B.S)}}{\le}\ \sum\frac{a^2}{bc} \)

\( \sum\frac{a^2}{bc}=\frac{a^3+b^3+c^3}{abc}=\frac{2p(p^2-6Rr-3r^2)}{4RS}=\frac{p^2-6Rr-3r^2}{2Rr}\ \stackrel{\mbox{(Gerretsen)}}{\le}\ \frac{4R^2+4Rr+3r^2-6Rr-3r^2}{2Rr}=\frac{2R-r}{r} \)

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\( \sum\frac{1}{a(b+c)}\ \stackrel{\mbox{(C.B.S.)}}{\ge}\ \frac{9}{2(ab+bc+ca)}=\frac{9}{2\underline{\underline{p^2}}+2r^2+8Rr}\ \stackrel{\mbox{(Gerretsen)}}{\ge}\ \frac{9}{2(\underline{\underline{4R^2+4Rr+3r^2}})+2r^2+8Rr}=\frac{9}{8(R+r)^2} \)