Question

demonstrati ca daca a,b∈( 0; +∞) sunt adevrate inegalitatile:

(a+b)/2≤(a²+b²)/a+b;

1/(a+b)≤1/4(1/a+1/b)

Answer 1

$\dfrac{(a+b)}{2}\leq \dfrac{a^2+b^2}{a+b}\\(a+b)^2\leq 2(a^2+b^2)\\a^2+b^2+2ab\leq 2a^2+2b^2\\a^2+b^2-2ab\geq 0\\(a-b)^2 \geq 0(A)\\\\\text{Putem aplica inegalitatea Titu Andreescu:}\\\dfrac{1}{a}+\dfrac{1}{b}\geq \dfrac{ (1+1)^2}{a+b}\\\dfrac{1}{a}+\dfrac{1}{b}\geq \dfrac{4}{a+b}|\cdot \dfrac{1}{4}\\\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\geq \dfrac{1}{a+b}$