Question

Aratati ca : 4$x^{2}$+1/4$x^{2}$> sau = cu 2
REPEDEEEE VA ROOOOG

Answer 1

(2x)²-2 *2x*1/(2x)+(1/2x)²≥0

(2x-1/2x)²≥0 adevarat∀x∈R

Answer 2

[tex]\text{Folosim inegalitatea mediilor:}\\ \boxed{\dfrac{a+b}{2}\geq \sqrt{a\cdot b}},\text{ unde a,b}\ \textgreater \ 0\\ \text{Atunci:}\\ \dfrac{4x^2+\frac{1}{4\cdot x^2}}{2}\geq \sqrt{4x^2\cdot \dfrac{1}{4x^2}}\\ \dfrac{4x^2+\frac{1}{4x^2}}{2}\geq 1\Rightarrow 4x^2+\dfrac{1}{4x^2}\geq 2[/tex]