Răspuns:
[tex]\sqrt{ab} \leq \frac{a+b}{2}~ |^2 ~~ab\leq \frac{(a+b)^{2}}{4} ~|*4~~4ab\leq a^{2}+2ab+b^{2}~~0\leq a^{2}+2ab+b^{2}-4ab~\\~0\leq a^{2}-2ab+b^{2}~~0\leq (a-b)^{2}~adevarat.\\Verificam~daca~\frac{2ab}{a+b} \leq \sqrt{ab} ~~|^2~~\frac{4(ab)^{2}}{(a+b)^{2}}\leq ab~~|*\frac{(a+b)^{2}}{ab} ~~4ab\leq (a+b)^{2}~~\\4ab\leq a^{2}+2ab+b^{2}~~0\leq a^{2}+2ab+b^{2}-4ab~~0\leq a^{2}-2ab+b^{2}~adevarat.\\[/tex]
Explicație pas cu pas:
Deci 2ab/(a+b)≤√(ab)≤(a+b)/2
Raspunsul e in imagine..
