[tex]\sqrt{a^2+b^2+c^2}= \sqrt{3} \Rightarrow a^2+b^2+c^2=3. \\ \\ Din~inegalitatea~(a+b+c)^2 \leq 3(a^2+b^2+c^2)=9,~rezulta \\ \\ a+b+c \leq 3. \\ \\ Din~inegalitatea~MA \leq MP~avem: \\ \\ \frac{\sqrt{a+b+1}+ \sqrt{b+c+1}+ \sqrt{a+c+1}}{3} \leq \sqrt{ \frac{2(a+b+c)+3}{3} } \leq \sqrt{ \frac{2 \cdot 3+3}{3} }= \sqrt{3}. \\ \\ Deci~\sqrt{a+b+1}+ \sqrt{b+c+1}+ \sqrt{a+c+1} \leq 3 \sqrt{3} . \\ \\ [/tex]
[tex]Tinand~cont~si~de~ipoteza,~rezulta~ca \\ \\ \sqrt{a+b+1}+ \sqrt{b+c+1}+ \sqrt{a+c+1} =3 \sqrt{3}. \\ \\ (aici~am~folosit~faptul~ca~x \geq y~si~x \leq y~implica~x=y) \\ \\ Acum,~tinand~cont~ca~egalitatea~in~fiecare~dintre~ \\ \\ inegalitatile~aplicate~are~loc ~ \Leftrightarrow a=b=c~si~ca~egalitatea \\ \\ trebuie~sa~aiba~loc ,~rezulta ~a=b=c \Rightarrow ~cub~de~latura~1 \Rightarrow \\ \\ \Rightarrow volum=1. [/tex]
*MA=media aritmetica // MP=media patratica
*Daca mi-ai fi citit mesajele, ai fi stiut de ce ti-am sters intrebarile.
