Question

sa se demonstreze ca daca a, b apartin R+ , atunci sunt adevarate urmatoarele inegalitati:

Answer 1

Răspuns:

a)

$a+b\ge 2\sqrt{ab}\Leftrightarrow (\sqrt{a}-\sqrt{b})^2\ge 0$

b)

$\dfrac{ab}{a+b}\le\dfrac{a+b}{4}\Leftrightarrow (a+b)^2\ge 4ab\Leftrightarrow a^2+2ab+b^2\ge 4ab\Leftrightarrow(a-b)^2\ge 0$

c)

$\dfrac{4}{a+b}\le\dfrac{1}{a}+\dfrac{1}{b}\Leftrightarrow\dfrac{4}{a+b}\le\dfrac{a+b}{ab}\Leftrightarrow\dfrac{ab}{a+b}\le \dfrac{a+b}{4}\Leftrightarrow b)$

d)

$\dfrac{a^2+b^2}{a+b}\ge\dfrac{a+b}{2}\Leftrightarrow 2a^2+2b^2\ge a^2+2ab+b^2\Leftrightarrow(a-b)^2\ge 0$

e)

$\dfrac{a+b}{a^2+b^2}\le\dfrac{a+b}{2ab}\Leftrightarrow a^2+b^2\ge 2ab\Leftrightarrow(a-b)^2\ge 0$

f)

$\dfrac{a}{b}+\dfrac{b}{a}\ge 2\Leftrightarrow a^2+b^2\ge 2ab\Leftrightarrow (a-b)^2\ge 0$

Explicație pas cu pas: