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Claim: For $x,y,z\in [-1,1]$, $xy+yz+zx\geq-1$. Proof: Let $x=sin\theta_1$, $y=sin\theta_2$, $z=sin\theta_3$. Therefore, $f=xy+yz+zx=sin\theta_1 sin\theta_2+sin\theta_2sin\theta_3+sin\theta_3sin\theta_1$. The minimum of $f$ happens when $\frac{\partial f}{\partial \theta_1}=cos\theta_1(sin\theta_2+sin\theta_3)=0$, and so on. Let us assume $\cos\theta_1\neq 0$. Therefore, $sin\theta_2+sin\theta_3=0$. This yields $f=sin\theta_1(sin\theta_2+sin\theta_3)+sin\theta_2sin\theta_3=-sin^2\theta_2\geq -1$. Which proves the inequality as long as there is some $i\in\{1,2,3\}$ for which $\cos\theta_i\neq0$. Now assume that for all $i\in\{1,2,3\}$, $\cos\theta_i=0$. In this case $\{x,y,z\}=\{1,1,1\},\{1,1,-1\},\{1,-1,-1\},\,or\,\{-1,-1,-1\}$, for which the inequality $f\geq-1$ still holds.
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Fri, 29 Dec 2017 07:10 GMT