For $x,y,z\in R$ and $x^2+xy+y^2=1$; $y^2+yz+z^2=16$

Prove: $xy+yz+zx\leq \frac{16}{3}$

  • $\begingroup$ Can we use Lagrange Multiplier Method? $\endgroup$ – gaoxinge Jan 5 '14 at 4:53
  • $\begingroup$ Using the Rotation of axes(en.wikipedia.org/wiki/Rotation_of_axes), $\displaystyle x'=\frac{x+y}{\sqrt2},y'=\frac{y-x}{\sqrt2}\iff x=\frac{x'-y'}{\sqrt2},y=\frac{x'+y'}{\sqrt2} $ $$x^2+xy+y^2=1\implies 3x'^2+y'^2=2\implies x'=\sqrt{\frac23}\cos\phi,y'=\sqrt{\frac13}\sin\phi$$ $\displaystyle\implies x=\frac{\sqrt2\cos\phi-\sin\phi}{\sqrt6},y=\frac{\sqrt2\cos\phi+\sin\phi}{\sqrt6} $ Similarly, $\displaystyle\frac y4=\frac{\sqrt2\cos\psi-\sin\psi}{\sqrt6},\frac z4=\frac{\sqrt2\cos\psi+\sin\psi}{\sqrt6} $ $\endgroup$ – lab bhattacharjee Jan 5 '14 at 5:23

$16=[(\frac x2+y)^2+\frac 34x^2][(\frac z2+y)^2+\frac 34z^2]$

Using the Cauchy-Schwarz inequality, we get

$16\ge \frac 34[(\frac x2+y)z+x(y+\frac z2)]^2$ $\Rightarrow xy+yz+zx\le \frac 8{\sqrt 3}<\frac {16}3$


Consider a triangle $\triangle$ABC, where AB=$1$ and BC=$4$, and a point P inside with $\angle$APB=$\angle$BPC=$120^\circ$. Thus we let AP=$x$, BP=$y$, CP=$z$, and $\frac{\sqrt{3}}{4}(xy+yz+xz)$ is the area of $\triangle$ABC, which is smaller or equal to $2$. So $$xy+yz+xz\leq \frac{8}{\sqrt{3}}<\frac{16}{3}$$

When some of the $x,y,z$ is negative, just reverse the direction of corresponding AP, BP or CP, and the proof remians the same.


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