[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]
