For $x,y,z\in R$ and $x^2+xy+y^2=1$; $y^2+yz+z^2=16$
Prove: $xy+yz+zx\leq \frac{16}{3}$
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Sign up to join this communityFor $x,y,z\in R$ and $x^2+xy+y^2=1$; $y^2+yz+z^2=16$
Prove: $xy+yz+zx\leq \frac{16}{3}$
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.
$16=[(\frac x2+y)^2+\frac 34x^2][(\frac z2+y)^2+\frac 34z^2]$
Using the Cauchy-Schwarz inequality, we get
$16\ge \frac 34[(\frac x2+y)z+x(y+\frac z2)]^2$ $\Rightarrow xy+yz+zx\le \frac 8{\sqrt 3}<\frac {16}3$