Question

Arătaţi că dacă x y, sunt numere reale strict pozitive astfel încât x *y =16 , atunci :
$\sqrt{2x+32y} + \sqrt{2y+32x} \geq 20$

Answer 1

$Avem:~x+y \geq 2 \sqrt{xy}=8. \\ \\Din~ \underline{inegalitatea ~lui~Minkovski},~avem: \\ \\ \sqrt{2x+32y}+ \sqrt{2y+32x}= \\ \\ = \sqrt{( \sqrt{2x})^2+( \sqrt{32y})^2} + \sqrt{( \sqrt{32x})^2+( \sqrt{2y})^2} \geq \\ \\ \geq \sqrt{( \sqrt{2x}+ \sqrt{32x})^2+ ( \sqrt{2y}+ \sqrt{32y})^2}=\\ \\ = \sqrt{2x+2 \cdot \sqrt{64x^2}+32x+2y+2 \sqrt{64y^2}+32y}= \\ \\ = \sqrt{50(x+y)} \geq \sqrt{50 \cdot 8}= \sqrt{400}=20,~q.e.d.$

$Egalitatea~are~loc~(printre~altele)~cand~x=y=4,~si~totusi \\ \\ pentru~x=y=4~nu~se~obtine~egalitate~:))~(Deci~inegalitatea~este \\ \\ stricta~(\ \textgreater \ 20)).$

$\underline{ INEGALITATEA~LUI~MINKOVSKI} \\ \\ \sqrt{x_1^2+x_2^2+...+x_n^2}+ \sqrt{y_1^2+y_2^2+...+y_n^2} \geq \\ \\ \geq \sqrt{(x_1+y_1)^2+(x_2+y_2)^2+...+(x_n+y_n)^2}, \forall~n \in N,~x_i,y_1 \in R~\forall~ \\ \\i \in \{1,2,...,n \}~(i= \overline{1,n}-scrierea~prescurtata).~Egalitate~ \Leftrightarrow \\ \\ \Leftrightarrow \frac{x_1}{y_1}=... \frac{x_n}{y_n} ~(daca~"y"-ii~sunt~nenuli).~\\ \\ .$