Inegalitate cu a,b,c>0.

Mesaj de **mihai miculita** » Mie Iun 10, 2015 1:02 pm

Aratati ca: $\dfrac{a+b}{a^2+b^2}+\dfrac{a+c}{a^2+c^2}+\dfrac{b+c}{b^2+c^2}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c};\,(\forall)a,b,c>0.$ (Regatele din Moskova-2015)

math111 · Mesaj de **math111** » Mie Iun 10, 2015 2:23 pm

mihai miculita scrie:Aratati ca: $\dfrac{a+b}{a^2+b^2}+\dfrac{a+c}{a^2+c^2}+\dfrac{b+c}{b^2+c^2}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c};\,(\forall)a,b,c>0.$ (Regatele din Moskova-2015)

$\dfrac{a+b}{a^2+b^2} \le \dfrac{1}{2} \left(\dfrac{1}{a}+\dfrac{1}{b}\right)$ si analoagele.

MathTime

Inegalitate cu a,b,c>0.

Inegalitate cu a,b,c>0.

Re: Inegalitate cu a,b,c>0.