O inegalitate

[DrAGos Calinescu]

Fie \( x,y,z\in\mathbb{R}_+^* \). Demonstrati urmatoarea inegalitate:
\( \frac{x}{\sqrt{y^2+yz+z^2}}+\frac{y}{\sqrt{x^2+xz+z^2}}+\frac{z}{\sqrt{x^2+xy+y^2}}\ge\sqrt{3} \)

[Claudiu Mindrila]

Cu inegalitatea Holder, avem:

\( \left(\sum_{cyc}\frac{x}{\sqrt{y^{2}+yz+z^{2}}}\right)\left(\sum_{cyc}\frac{x}{\sqrt{y^{2}+yz+z^{2}}}\right)\left[\sum_{cyc}x\left(y^{2}+yz+z^{2}\right)\right]\ge\left(x+y+z\right)^{3} \)

adica \( \left(\sum_{cyc}\frac{x}{\sqrt{y^{2}+yz+z^{2}}}\right)^{2}\ge\frac{\left(x+y+z\right)^{3}}{\sum_{cyc}x\left(y^{2}+yz+z^{2}\right)} \).

Mai avem de aratat ca \( \frac{\left(x+y+z\right)^{3}}{\sum_{cyc}x\left(y^{2}+yz+z^{2}\right)}\ge3\Longleftrightarrow\left(x+y+z\right)^{3}\ge3\sum_{cyc}xy\left(x+y\right)+9xyz \),

adica \( \left(x+y+z\right)^{3}\ge3\sum_{cyc}xy\left(x+y\right)+9xyz\Longleftrightarrow\sum_{cyc}x^{3}+3\sum_{cyc}xy\left(x+y\right)+6xyz\ge3\sum_{cyc}xy\left(x+y\right)+9xyz \)

care se reduce la \( x^{3}+y^{3}+z^{3}\ge3xyz \), inegalitate evident adevarata.