Question

exercitiul 55 va rog mult

Answer 1

$Cerinta~este~echivalenta~cu: \\ \\ 4x^2+4y^2+4z^2+4t^2-4x+4y-4z+4t+1 \in [0;148]. \\ \\ Notand~expresia~de~mai~sus~cu~E,~avem~de~demonstrat~ca~ \\ \\ E \in[0;148]. \\ \\ E=(4x^2-4+1)+(4y^2+4y+1)+(4z^2-4z+1)+(4t^2+4t+1)= \\ \\ =(2x-1)^2+(2y+1)^2+(2z-1)^2+(2t+1)^2 \geq 0+0+0+0=0, \\ \\ Deci~E \geq 0.~(egalitate~pentru~x=y=z=t= \frac{1}{2})~~~~~(1)$

$-2 \leq x \leq 1 \Rightarrow -5 \leq 2x-1 \leq 1 \Rightarrow (2x-1)^2 \leq 25. \\ \\ -1 \leq y \leq 2 \Rightarrow -1 \leq 2y+1 \leq 5 \Rightarrow (2y-1)^2 \leq 25. \\ \\ -3 \leq z \leq 4 \Rightarrow -7 \leq 2z-1 \leq 7 \Rightarrow (2z-1)^2 \leq 49. \\ \\ -4 \leq t \leq 3 \Rightarrow -7 \leq 2t+1 \leq 7 \Rightarrow (2t+1)^2 \leq 49.$

$Deci~E \leq 25+25+49+49=148.~(egalitatea~poate~avea~loc) \\ \\ Deci~E \in [0;148],~de~unde~rezulta~ca~ \frac{E}{4} \in[0;37]. \\ \\ \frac{E}{4} ~fiind~chiar~expresia~din~problema.$