For which values of x, y does the equality $$x^2y^2+x^2+y^2+4 \leq 6xy$$ hold ? Please could you help with this problem as I am having trouble getting started.
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The inequality is true for all $x,y$ since the inequality can be rewritten as follows $0\leq (x^2y^2-4xy+4)+(x^2+y^2-2xy)=(xy-2)^2+(x-y)^2$
user9077
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As this is symmetric in $x$ and $y$, use the elementary symmetric functions $s=x+y$ and $p=xy$. The inequality translates into $p^2+(s^2-2p)+4\ge 6p$. Now $x$ and $y$ are the (real) roots of the quadratic equation $t^2-st+p=0$, and $s^2-4p\ge 0$, so that $$p^2+(s^2-2p)+4\ge p^2+2p+4.$$ The latter expression is $\ge6p$ since $p^2-4p+4=(p-2)^2\ge0 $.
Jean Marie
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Bernard
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