Question

$Daca~x,y,z \geq 0~si~x+y+z=3,aflati ~maximul~si~minimul \\ expresiei~ \sqrt{x(y+3)} + \sqrt{y(z+3)} + \sqrt{z(x+3)} .$

Answer 1

$\displaystyle 2E= \sqrt{4x(y+3)}+ \sqrt{4y(z+3)}+ \sqrt{4z(x+3)} \\ \\ Din~inegalitatea~mediilor~(AM-GM),~avem: \\ \\ \sqrt{4x(y+3)} \leq \frac{4x+y+3}{2}~si~analoagele. \\ \\ Deci~2E \leq \frac{4x+y+3}{2}+ \frac{4y+z+3}{2}+ \frac{4z+x+3}{2}=12. \\ \\ Deci~E \le 6,~iar~egalitatea~are~loc~pentru~x=y=z=1,~ceea \\ \\ ce~inseamna~ca~maximul~este~6.$

$\displaystyle E^2=x(y+3)+y(z+3)+z(x+3)+2 \sqrt{xy(y+3)(z+3)}+ \\ \\ +2 \sqrt{xz(x+3)(y+3)}+2 \sqrt{yz(x+3)(z+3)}= \\ \\ =9+xy+yz+xz+2(suma~de~radicali) \geq 9+0+0=9. \\ \\ Deci~E \geq 3.~Observam~ca~egalitatea~poate~avea~loc~(de~exemplu \\ \\ pentru~x=0,~y=0,~z=3),~deci~minimul~este~3.$