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Prove that $x^2+y^2+z^2+t(xy+yz+zx) \geq 0$ for any $x,y,z \in \mathbb{R}$ and any $t \in [-1,2].$

One try:

for $t=-1$: $x^2+y^2+z^2-xy-yz-zx \geq 0$ is true .

for $t=2$: $x^2+y^2+z^2+2(xy+yz+zx) \geq 0$ is true

also for $t=0$.

But how can prove for $t \in (-1,0)$ and $t \in (0,2)$.

Also I don't know if what I did is the right step.

Thanks :)

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1 Answer 1

up vote 5 down vote accepted

Note that

\begin{eqnarray} x^2+y^2+z^2+t(xy+yz+zx)&=&\frac{2-t}{3}(x^2+y^2+z^2-xy-yz-zx)\\ &&+\frac{1+t}{3}(x^2+y^2+z^2+2xy+2yz+2yz) \, . \end{eqnarray}

Your previous work shows that both terms on the right-hand side are always non-negative for $t \in [-1, 2]$; thus the left-hand side is as well.

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More generally, if $u+vt_0\geq 0$ and $u+vt_1\geq 0$ with $t_1<t_2$ then $u+vt\geq 0$ for all $t\in[t_0,t_1]$. –  Thomas Andrews Feb 27 '13 at 18:12
    
I feel stupid for not seeing this answer - got buried in algebraic manipulation, when it just amounts to "the half-plane is convex." :) –  Thomas Andrews Feb 27 '13 at 18:19
    
it's a nice solution - and I tried to combine the relations, but nothing. Thanks again :) –  Iuli Feb 27 '13 at 18:43
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