Răspuns:
este dat ca a,b>0, deci putem inmulti si ridica la patrat partile unei inegalitati si semnul inegalitatii nu se schimba...
am mai folosit ca daca x≤y, atunci y≥x
Explicație pas cu pas:
[tex]d)~a+\frac{1}{a} \geq 2,~|*a,~a^{2}+1\geq 2a,~a^{2}-2a+1\geq 0,~(a-1)^{2}\geq 0~este~adevarat\\f)~\frac{a+b}{2} \leq \sqrt{\frac{a^{2}+b^{2}}{2}},~ridicam~la~patrat~ambele~parti~nenegative\\ (\frac{a+b}{2})^{2}\leq (\sqrt{\frac{a^{2}+b^{2}}{2}})^{2},~\frac{(a+b)^{2}}{4}\leq \frac{a^{2}+b^{2}}{2}},~|*4,~(a+b)^{2}\leq 2*(a^{2}+b^{2}),~a^{2}+b^{2}+2ab\leq 2a^{2}+2b^{2},~2ab\leq a^{2}+b^{2},~sau~a^{2}+b^{2}\geq 2ab,~a^{2}+b^{2}-2ab\geq 0,~(a-b)^{2}\geq 0,~este~adevarat[/tex]
