Aratati ca :
$(a+2b+\dfrac{2}{a+1})(b+2a+\dfrac{2}{b+1}) \ge 16$
pentru toate numerele reale pozitive $a$ si $b$, cu $ab\ge 1$.
JBMO 2013 pr 3
-
Stefan Dominte
- Mesaje: 61
- Membru din: Mar Mai 15, 2012 8:32 pm
- Localitate: Iasi
Re: JBMO 2013 pr 3
"We must know, we will know" - David Hilbert
Re: JBMO 2013 pr 3
Scriem fiecare paranteza astfel
$(\frac{a+1}{2}+\frac{2}{a+1}+\frac{a-1}{2}+2b)(\frac{b+1}{2}+\frac{2}{b+1}+\frac{b-1}{2}+2a)$ $\ge (2+\frac{a-1}{2}+2b)(2+\frac{b-1}{2}+2a)$.
Ramane de demonstrat
$(3+a+4b)(3+b+4a) \ge 64$. Din CBS,
$(3+a+4b)(3+b+4a) \ge (3+\sqrt{ab}+4\sqrt{ab})^2 \ge (3+1+4)^2=64$. Egalitate pentru $a=b=1$.
$(\frac{a+1}{2}+\frac{2}{a+1}+\frac{a-1}{2}+2b)(\frac{b+1}{2}+\frac{2}{b+1}+\frac{b-1}{2}+2a)$ $\ge (2+\frac{a-1}{2}+2b)(2+\frac{b-1}{2}+2a)$.
Ramane de demonstrat
$(3+a+4b)(3+b+4a) \ge 64$. Din CBS,
$(3+a+4b)(3+b+4a) \ge (3+\sqrt{ab}+4\sqrt{ab})^2 \ge (3+1+4)^2=64$. Egalitate pentru $a=b=1$.
Catană Adrian,
Elev la CNIV, Targoviste, clasa a X-a
Elev la CNIV, Targoviste, clasa a X-a