MathTime

Lema : Fie $\triangle ABC$ un triunghi cu $\angle B=30^{\circ}$ si $\angle C=20^{\circ}$. Fie un punct $D\in(BC)$. Aratati ca $BD=AC\Leftrightarrow DA=DC$. solutia lemei o gasiti aici - http://forum.gil.ro/viewtopic.php?f=20&t=154 Aplicatii : 1. Fie un triunghi $\triangle ABC$ cu $m(\angle B)=1...

Solutie (Bogdan Mihai, profesor Constanta): $\dfrac{a}{1+2b^3}=a-\dfrac{2ab^3}{1+2b^3}=a-\dfrac{2ab^3}{1+b^3+b^3} \geqslant a - \dfrac{2ab^3}{3 \sqrt[3]{b^6}}=a-\dfrac{2ab^2}{3b^2}=a-\dfrac{2ab}{3}$ $\sum \dfrac{1}{2b^3} \geqslant a+b+c+d-\dfrac{2}{3}(ab+bc+cd+da)=$ $=\dfrac{3(a+b+c+d)-2(ab+bc+cd+d...

Fie $a,b,c,d$ numere reale nenegative astfel incat $a+b+c+d=3$. Demonstrati ca

$\frac{a}{1+2b^{3}}+\frac{b}{1+2c^{3}}+\frac{c}{1+2d^{3}}+\frac{d}{1+2a^{3}}\geq\frac{a^{2}+b^{2}+c^{2}+d^{2}}{3}.$

Cand are loc egalitatea?

Marius Stanean

Triunghiul $ABC$ cu $AC=BC$ si $ACB=96^{\circ}$. Fie $D$ un punct in interiorul $\triangle ABC$ astfel incat $m(\widehat{DAB})=18^{\circ}$ si $m(\widehat{DBA})=30^{\circ}$. Calculati $m(\widehat{ACD})$.

Sa se demonstreze ca daca $a,b,c>0$ si $abc=1$, atunci
$$a^{3}+b^{3}+c^{3}+\frac{2ab}{a^{2}+b^{2}}+\frac{2bc}{b^{2}+c^{2}}+\frac{2ca}{c^{2}+a^{2}}\geq 6.$$

Succes

La multi ani cu multa sanatate si succes in continuare!

La multi ani cu multa sanatate si succes in continuare!

La multi ani si multa sanatate!!!

Se considera triunghiul $\triangle{ABC}$, cu $AB=AC$ si $m(\angle{BAC})=90^\circ$. Fie $D \in (BC)$
astfel incat $AD \perp BC$. Bisectoarea unghiului $\angle{ABC}$ interseteaza dreapta $AD$ in punctul $I$.
Aratati ca $BA+AI=BC$