Question

Fie a, b, x, y numere reale. Atunci au loc inegalitatile:
a) $(ax+by) ^{2} \leq (a^{2}+b^{2})(x^{2}+y^{2})$
(inegalitateaCauchy-Buniakowski-Schwartz)
b)$\sqrt{(a+b)^{2}+(x+y)^{2} } \leq \sqrt{a^{2}+x^{2}}+ \sqrt{x^{2}+y^{2}}$
(inegalitatea lui Minkowshi).

Pls help ;(

Answer 1

a)(ax+by)²≤(a²+b²)(x²+y²)
a²x²+2axby+b²y²≤a²x²+b²x²+b²y²+a²y²
a²x² si b²y² se reducc si vom avea:
2axby≤b²x²+a²y²
Trecem pe 2axby in partea cealalta:
b²x²+a²y²-2axby≥0
(bx-ay)²≥0 (A)
b)[tex] \sqrt{(a+b)^2+(x+y)^2} \leq \sqrt{a^2+x^2}+\sqrt{b^2+y^2}\\ Ridicam\ fiecare\ membru\ la\ patrat.\\ (a+b)^2+(x+y)^2 \leq a^2+x^2+b^2+y^2+2\sqrt{(a^2+x^2)(b^2+y^2)\\ [/tex][tex]a^2+2ab+b^2+x^2+2xy+y^2 \leq\\ a^2+x^2+b^2+y^2+ 2\sqrt{(a^2+x^2)(b^2+y^2) \\ [/tex]
[tex]2ab+2xy \leq 2\sqrt{(a^2+x^2)(b^2+y^2)}\\ ab+xy \leq \sqrt{(a^2+x^2)(b^2+y^2)}\\ Ridicam\ iarasi\ totul\ la\ patrat:\\ (ab+xy)^2 \leq (a^2+x^2)(b^2+y^2)\\ Si\ acum\ se\ demonstreaz\ egalitatea\ Cauchy-Schwartz\ dupa\\ exercitiul\ de\ mai\ sus\. [/tex]