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