ملتقى الفيزيائيين العرب - عرض مشاركة واحدة

Weierstrass-Casorati · #5 23-08-2010, 18:15

$\\ {\color{red} (1)}\, \frac{2}{a+b}+\frac{2}{c+a}+\frac{2}{b+c}\geq \frac{(3\sqrt{2})^2}{2(a+b+c)}=\frac{9}{a+b+c}\\ \\ \\ {\color{red} (2)}\,a^2+b^2\geq \frac{(a+b)^2}{2}\Rightarrow (a^2+b^2)^2 \geq\frac{(a+b)^4}{4}\\ \\ a^4+b^4\geq\frac{(a^2+b^2)^2}{2}\\ \\ a^4+b^4\geq \frac{(a+b)^4}{8}\\ \\ 8\left (a^4+b^4 \right )\geq \left ( a+b\right )^4 \\ \\ {\color{red} (4)}\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{1}{2}(a+b+c)...(1)\\ \\ \frac{a^2}{c+a}+\frac{b^2}{a+b}+\frac{c^2}{b+c}\geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{1}{2}(a+b+c)...(2)\\ \\ (1)+(2)\Rightarrow \frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\geq a+b+c$