Question

với x, y, z > 0. CMR :

\(\dfrac{x^2}{\sqrt{8x^2+3y^2+14xy}}+\dfrac{y^2}{\sqrt{8y^2+3z^2+14yz}}+\dfrac{z^2}{\sqrt{8z^2+3x^2+14zx}}\ge\dfrac{x+y+z}{5}\)

Answer 1

\(\Sigma\dfrac{x^2}{\sqrt{8x^2+3y^2+14xy}}=\Sigma\dfrac{x^2}{\sqrt{\left(4x+y\right)\left(2x+3y\right)}}\ge\dfrac{2\left(x+y+z\right)^2}{4x+y+2x+3y+4y+z+2y+3z+4z+x+2z+3x}=\dfrac{\left(x+y+z\right)^2}{5\left(x+y+z\right)}=\dfrac{x+y+z}{5}\left(đpcm\right)\)

Answer 2

\(\sqrt{8x^2+3y^2+14xy}=\sqrt{8x^2+2xy+3y^2+12xy}\)

\(\le\sqrt{9x^2+4y^2+12xy}\left(vì\left(2xy\le x^2+y^2\right)\right)\)

\(\le3x+2y\)

\(\Rightarrow\dfrac{x^2}{\sqrt{8x^2+3y^2+14xy}}\ge\dfrac{x^2}{3x+2y}\)

\(\Rightarrow\text{Σ}\dfrac{x^2}{\sqrt{8x^2+3y^2+14xy}}\ge\dfrac{\left(x+y+z\right)^2}{5\left(x+y+z\right)}=\dfrac{x+y+z}{5}\left(BĐT\text{Svácxơ}\right)\)