Question

Sa se determine valoarea minima a numarului N= $\sqrt{ x^{2} +6x+25}$ + $\sqrt{2 y^{2}+8y+24} }$.
Va rooog ajutati-ma.

Answer 1

$N=\sqrt{x^2+2\cdot 3x+3^2+4^2}+\sqrt{2(y^2+4\cdot y+12)}=\\\\=\sqrt{x^2+2\cdot 3x+3^2+4^2}+\sqrt{2(y^2+2\cdot 2y+2^2+8)}=\\\\=\sqrt{(x+3)^2+4^2}+\sqrt{2[(y+2)^2+8]};\\\\(x+3)^2\geqslant 0\Rightarrow (x+3)^2+4^2\geqslant4^2\Rightarrow \sqrt{(x+3)^2+4^2}\geqslant 4\;(1);\\\\(y+2)^2\geqslant0\Rightarrow (y+2)^2+8\geqslant8\Rightarrow 2[(y+2)^2+8]\geqslant 16\Rightarrow\\\\\Rightarrow\sqrt{2[(y+2)^2+8]}\geqslant 4\;(2).\ Dac\breve{a}\;aduni\;rela\c{t}iile\ (1)\ si\;(2)\\\\membru\;cu\ membru\ ob\c{t}ii\;c\breve{a}\;N_{min}=8.$

Green eyes.